MathsCBSE Class 10 · Chapter 10

Circles Class 10: Tangents, Theorems and Solved Problems

Q: Which proofs must I learn for Circles in Class 10?

You must learn two proofs: the tangent at any point is perpendicular to the radius through the point of contact, and the tangents from an external point are equal in length. Both are named in the syllabus as proofs, so write them with Given, To prove, Construction and full reasons.

Q: Are constructions of tangents in the 2026-27 syllabus?

No. Constructions of tangents were removed in the NCERT rationalisation and are not in the 2026-27 syllabus.

Q: Do I need to draw the figure in a proof?

Yes. A neat labelled figure is expected, and in a proof it carries part of the marks along with Given and To prove. Mark the right angles at the points of contact on the figure.

Q: How do I know which side is the hypotenuse in tangent problems?

The right angle is always at the point of contact, between the radius and the tangent. The side opposite it, from the centre to the external point, is the hypotenuse. So the square of the tangent length equals the square of that distance minus the square of the radius.

Q: Can I directly quote results like the angle between tangents being supplementary to the angle at the centre?

In an MCQ or assertion-reason question you can use it directly. In a 2-, 3- or 5-mark answer it is safer to show the two lines of working from the quadrilateral formed by the centre, the two points of contact and the external point, because the examiner looks for the reasons.

Updated 13 October 2026 · Fact-checked

Circles in Class 10 is about tangents. You learn two theorems: a tangent is perpendicular to the radius at the point of contact, and the two tangents from an outside point are equal in length. The board asks you to prove these, use them in angle and length problems, and apply them to triangles and quadrilaterals around a circle.

What Circles is about

This chapter studies one special line: the tangent, a line that touches a circle at exactly one point. Everything rests on two theorems. First, the tangent at any point is perpendicular to the radius through that point. Second, the two tangents drawn from an external point are equal in length. Once you can prove both and use them, you can solve almost every question in the chapter.

The chapter links closely to others. The right angle between tangent and radius turns many problems into Pythagoras questions from Triangles. Congruence (RHS) from Class 9 drives the proof of equal tangents. Trigonometric ratios help when an angle between tangents is given, and the distance formula from Coordinate Geometry tells you whether a point is inside, on or outside a circle.

Board questions come in a few clear shapes: angle-chasing MCQs with two tangents, assertion-reason items on the properties, 2- and 3-mark length problems, full proofs of the two theorems or results built on them (often in Section D), problems on triangles and quadrilaterals drawn around a circle, and real-life case studies such as wheels, belts or bangles.

Weightage: This chapter is part of Unit IV, Geometry, which carries 15 of the 80 theory marks together with Triangles.Source: CBSE Mathematics syllabus 2026-27, Unit IV Geometry

In real board papers: Standard, asked in 21 of the 21 board papers we hold (2023 to 2026), on average 7.2 marks per paper; Basic, asked in 21 of the 21 board papers we hold (2023 to 2026), on average 6.5 marks per paper. See the questions

Key concepts

Secant, tangent and the point of contact

NCERT section 10.2 Tangent to a Circle

A line and a circle can meet in three ways. If the line misses the circle, it is non-intersecting. If it cuts the circle at two points, it is a secant. If it meets the circle at exactly one point, it is a tangent, and that point is the point of contact.

Why does a tangent touch at only one point? Think of a secant through two points $A$ and $B$ on the circle. Slide $B$ towards $A$ . The chord $A B$ shrinks, and when $B$ reaches $A$ the secant becomes the tangent at $A$ . So a tangent is the limiting position of a secant.

A useful way to test a line: find the perpendicular distance $d$ from the centre to the line. If $d > r$ the line misses the circle, if $d = r$ it is a tangent, and if $d < r$ it is a secant. There is exactly one tangent at each point of a circle, so a circle has infinitely many tangents.

Worked example

A circle has centre $O$ and radius $5$ cm. $A B$ is a chord of length $8$ cm. A line parallel to $A B$ is a tangent to the circle. Find the possible distances between this tangent and the chord $A B$ .

Try it first, then show the solutionSolution

Draw $O M ⊥ A B$ . The perpendicular from the centre to a chord bisects it, so $A M = 4$ cm.
In right $△ O M A$ : $O M^{2} = O A^{2} - A M^{2} = 25 - 16 = 9$ , so $O M = 3$ cm.
A line parallel to $A B$ is also perpendicular to line $O M$ , so its distance from $O$ is measured along $O M$ .
It is a tangent exactly when its distance from $O$ equals the radius, $5$ cm.
Tangent on the same side of $O$ as $A B$ : distance from $A B = 5 - 3 = 2$ cm.
Tangent on the opposite side of $O$ : distance from $A B = 5 + 3 = 8$ cm.

Answer: The tangent is either $2$ cm or $8$ cm from the chord $A B$ .

Theorem 1: the tangent is perpendicular to the radius (proof)

NCERT section 10.2 Tangent to a Circle

Statement: The tangent at any point of a circle is perpendicular to the radius through the point of contact.

Proof idea. Let the tangent $X Y$ touch the circle with centre $O$ at $P$ . Take any other point $Q$ on $X Y$ . Since a tangent meets the circle only at $P$ , $Q$ lies outside the circle. So $O Q$ is longer than the radius: $O Q > O P$ . This is true for every point $Q$ on $X Y$ other than $P$ . So $O P$ is the shortest distance from $O$ to the line $X Y$ . Of all segments from a point to a line, the shortest is the perpendicular. Hence $O P ⊥ X Y$ .

In the board proof you must write the reason for $O Q > O P$ (Q is outside the circle) and the fact that the shortest segment is the perpendicular. Two consequences follow: the line through $P$ perpendicular to the tangent passes through the centre, and tangents at the ends of a diameter are parallel.

Worked example

$A B$ is a diameter of a circle with centre $O$ . Line $l$ is the tangent at $A$ and line $m$ is the tangent at $B$ . Prove that $l ∥ m$ .

Try it first, then show the solutionSolution

$O A$ is a radius and $l$ is the tangent at $A$ , so $O A ⊥ l$ (tangent is perpendicular to radius at point of contact).
So the angle between $l$ and line $A B$ at $A$ is $9 0^{\circ}$ .
Similarly $O B ⊥ m$ , so the angle between $m$ and line $A B$ at $B$ is $9 0^{\circ}$ .
Line $A B$ is a transversal to $l$ and $m$ , and the two angles on the same side of it sum to $9 0^{\circ} + 9 0^{\circ} = 18 0^{\circ}$ .
Co-interior angles are supplementary, so $l ∥ m$ .

Answer: Hence the tangents at the ends of a diameter are parallel. The distance between them is the diameter, $2 r$ .

Using the right angle: Pythagoras with tangents

NCERT section 10.2 Tangent to a Circle

Theorem 1 gives you a right angle every time a radius meets a tangent. Join the centre $O$ to the external point $P$ and to the point of contact $T$ . Then $△ O T P$ is right-angled at $T$ , with **hypotenuse $O P$ **. So $P T^{2} = O P^{2} - r^{2}$ .

This one triangle answers three kinds of question: find the tangent length, the radius, or the distance of $P$ from the centre, given the other two.

Watch for disguised data. The distance from $P$ to the circle is not $O P$ ; if line $P O$ meets the circle first at $A$ , then $O P = P A + r$ . With concentric circles, a chord of the bigger circle that touches the smaller one is perpendicular to the small radius at the contact point, so it is bisected there. Its length is $2 R^{2} - r^{2}$ . Always state the reason for the right angle before using Pythagoras.

Worked example

$P T$ is a tangent from an external point $P$ to a circle with centre $O$ , touching it at $T$ . The line $P O$ meets the circle at $A$ , the point nearer to $P$ . If $P T = 12$ cm and $P A = 8$ cm, find the radius of the circle.

Try it first, then show the solutionSolution

Let the radius be $r$ cm. Then $O T = O A = r$ and $O P = P A + A O = 8 + r$ .
$O T ⊥ P T$ (tangent is perpendicular to radius), so $△ O T P$ is right-angled at $T$ .
By Pythagoras: $O P^{2} = O T^{2} + P T^{2}$ , so $(r + 8)^{2} = r^{2} + 144$ .
$r^{2} + 16 r + 64 = r^{2} + 144$ , so $16 r = 80$ and $r = 5$ .
Check: $O P = 13$ and $1 3^{2} = 169 = 25 + 144$ . Correct.

Answer: The radius is $5$ cm.

Number of tangents from a point

NCERT section 10.3 Number of Tangents from a Point on a Circle

How many tangents you can draw depends on where the point is.

Inside the circle: none. Every line through an inside point cuts the circle twice, so every such line is a secant.
On the circle: exactly one. It is the line through the point perpendicular to the radius there.
Outside the circle: exactly two, one touching on each side of the line joining the point to the centre.

To decide where a point is, compare its distance from the centre with the radius: less than $r$ means inside, equal means on, greater means outside. When coordinates are given, use the distance formula. This idea often appears as a 1-mark MCQ or as the reason part of an assertion-reason question, so learn the three cases with their reasons, not just the numbers.

Worked example

A circle has centre at the origin and radius $5$ units. From which point can exactly two tangents be drawn to the circle?
(a) $(3, 4)$
(b) $(1, 2)$
(c) $(0, - 5)$
(d) $(4, 4)$

Try it first, then show the solutionSolution

Find each point's distance from the origin $O (0, 0)$ using $d = x^{2} + y^{2}$ .
$(3, 4)$ : $d = 9 + 16 = 5$ , on the circle, so one tangent.
$(1, 2)$ : $d = 5 < 5$ , inside, so no tangent.
$(0, - 5)$ : $d = 5$ , on the circle, so one tangent.
$(4, 4)$ : $d = 32 > 25 = 5$ , outside, so two tangents.

Answer: (d) $(4, 4)$

Theorem 2: tangents from an external point are equal (proof)

NCERT section 10.3 Number of Tangents from a Point on a Circle

Statement: The lengths of tangents drawn from an external point to a circle are equal.

Proof. Let $P A$ and $P B$ be tangents from $P$ to a circle with centre $O$ , touching it at $A$ and $B$ . Join $O A$ , $O B$ and $O P$ . In $△ O A P$ and $△ O B P$ :

$∠ O A P = ∠ O B P = 9 0^{\circ}$ (tangent is perpendicular to radius),
$O A = O B$ (radii),
$O P = O P$ (common hypotenuse).

So $△ O A P ≅ △ O B P$ by RHS, and $P A = P B$ by CPCT.

The same congruence gives more for free: $∠ A P O = ∠ B P O$ , so $O P$ bisects the angle between the tangents, and $∠ A O P = ∠ B O P$ . Also, since $P A = P B$ and $O A = O B$ , $O P$ is the perpendicular bisector of $A B$ . These extra facts are used constantly in angle and length problems.

Worked example

$P A$ and $P B$ are tangents from $P$ to a circle with centre $O$ . If $O P = 10$ cm and $∠ A P B = 6 0^{\circ}$ , find the radius of the circle and the length of the chord $A B$ .

Try it first, then show the solutionSolution

$△ O A P ≅ △ O B P$ (RHS), so $O P$ bisects $∠ A P B$ : $∠ O P A = 3 0^{\circ}$ .
$∠ O A P = 9 0^{\circ}$ (tangent is perpendicular to radius).
In right $△ O A P$ : $sin 3 0^{\circ} = \frac{O A}{O P}$ , so $O A = 10 \times \frac{1}{2} = 5$ cm.
$cos 3 0^{\circ} = \frac{P A}{O P}$ , so $P A = 10 \times \frac{3}{2} = 53$ cm.
$P A = P B$ (tangents from an external point) and $∠ A P B = 6 0^{\circ}$ , so $△ P A B$ is equilateral.
Hence $A B = P A = 53$ cm.

Answer: Radius $= 5$ cm and $A B = 53$ cm.

Angle between two tangents and the angle at the centre

NCERT section 10.3 Number of Tangents from a Point on a Circle

Join the centre $O$ to the points of contact $A$ and $B$ of tangents from $P$ . The quadrilateral $O A P B$ has $∠ O A P = ∠ O B P = 9 0^{\circ}$ . Its angles add to $36 0^{\circ}$ , so

$∠ A P B + ∠ A O B = 18 0^{\circ}$ .

The angle between the tangents and the angle subtended by $A B$ at the centre are supplementary.

Two more results come from isosceles triangles. $P A = P B$ , so $∠ P A B = ∠ P B A = 9 0^{\circ} - \frac{1}{2} ∠ A P B$ . And since $∠ O A B = 9 0^{\circ} - ∠ P A B$ , you get $∠ O A B = \frac{1}{2} ∠ A P B$ . In other words, the angle between the tangents is twice the angle between the chord $A B$ and a radius. You should be able to derive these in a few lines rather than quote them, because proofs of exactly this kind are asked.

Worked example

$P A$ and $P B$ are tangents from an external point $P$ to a circle with centre $O$ . If $∠ O A B = 2 8^{\circ}$ , find $∠ A P B$ .

Try it first, then show the solutionSolution

$O A = O B$ (radii), so $∠ O B A = ∠ O A B = 2 8^{\circ}$ .
In $△ O A B$ : $∠ A O B = 18 0^{\circ} - 2 8^{\circ} - 2 8^{\circ} = 12 4^{\circ}$ .
In quadrilateral $O A P B$ , $∠ O A P = ∠ O B P = 9 0^{\circ}$ (tangent is perpendicular to radius).
So $∠ A P B = 36 0^{\circ} - 9 0^{\circ} - 9 0^{\circ} - 12 4^{\circ} = 5 6^{\circ}$ .
Check: $∠ A P B = 2∠ O A B = 2 \times 2 8^{\circ} = 5 6^{\circ}$ . Correct.

Answer: $∠ A P B = 5 6^{\circ}$

Triangle drawn around a circle (incircle)

NCERT section 10.3 Number of Tangents from a Point on a Circle

If a circle touches all three sides of $△ A B C$ , at $D$ on $B C$ , $E$ on $C A$ and $F$ on $A B$ , then each vertex is an external point with two tangents. So

$A F = A E$ , $B F = B D$ , $C D = C E$ .

Call these $x$ , $y$ , $z$ . Then $x + y = A B$ , $y + z = B C$ , $z + x = C A$ . Adding, $x + y + z = s$ , the semi-perimeter. Hence $A F = s - B C$ , $B D = s - C A$ , $C E = s - A B$ : each tangent length is the semi-perimeter minus the opposite side.

For a right triangle with legs $a$ , $b$ and hypotenuse $c$ , the radii to the two legs and the legs themselves form a square at the right angle, which gives $r = \frac{a + b - c}{2}$ . Also, joining the centre to the vertices splits the triangle into three triangles of height $r$ , so area $= \frac{1}{2} r (a + b + c)$ .

Worked example

A circle is inscribed in $△ A B C$ touching $B C$ , $C A$ and $A B$ at $D$ , $E$ and $F$ respectively. If $A B = 12$ cm, $B C = 8$ cm and $C A = 10$ cm, find $A F$ , $B D$ and $C E$ .

Try it first, then show the solutionSolution

Tangents from an external point are equal: $A F = A E = x$ , $B F = B D = y$ , $C D = C E = z$ .
$x + y = 12$ , $y + z = 8$ , $z + x = 10$ .
Adding: $2 (x + y + z) = 30$ , so $x + y + z = 15$ .
$x = 15 - (y + z) = 15 - 8 = 7$ .
$y = 15 - (z + x) = 15 - 10 = 5$ .
$z = 15 - (x + y) = 15 - 12 = 3$ .
Check: $B D + D C = 5 + 3 = 8 = B C$ . Correct.

Answer: $A F = 7$ cm, $B D = 5$ cm, $C E = 3$ cm.

Quadrilateral drawn around a circle

NCERT section 10.3 Number of Tangents from a Point on a Circle

If a circle touches all four sides of quadrilateral $A B C D$ , at $P$ on $A B$ , $Q$ on $B C$ , $R$ on $C D$ and $S$ on $D A$ , then from each vertex the two tangents are equal: $A P = A S$ , $B P = B Q$ , $C R = C Q$ , $D R = D S$ .

Add the four equations side by side. The left side groups into $A B + C D$ and the right side into $A D + B C$ . So

$A B + C D = A D + B C$ .

The sums of opposite sides are equal. Two famous consequences: a parallelogram that circumscribes a circle must be a rhombus (opposite sides are equal, so $2 A B = 2 B C$ ), and a rectangle that circumscribes a circle must be a square.

If a vertex angle is $9 0^{\circ}$ , the two radii to the sides at that vertex and the two tangent segments form a square, so each tangent length from that vertex equals $r$ . Also, the opposite sides of such a quadrilateral subtend supplementary angles at the centre.

Worked example

Quadrilateral $A B C D$ circumscribes a circle with centre $O$ and radius $6$ cm, touching $A B$ , $B C$ , $C D$ , $D A$ at $P$ , $Q$ , $R$ , $S$ respectively. $∠ A = 9 0^{\circ}$ , $B C = 18$ cm, $B Q = 12$ cm and $C D = 9$ cm. Find $A B$ and $A D$ .

Try it first, then show the solutionSolution

$O P ⊥ A B$ and $O S ⊥ A D$ (tangent is perpendicular to radius), and $∠ A = 9 0^{\circ}$ .
So $A P O S$ has four right angles and $O P = O S = 6$ cm; it is a square, so $A P = A S = 6$ cm.
$B P = B Q = 12$ cm (tangents from $B$ ), so $A B = A P + P B = 6 + 12 = 18$ cm.
$C Q = B C - B Q = 18 - 12 = 6$ cm, and $C R = C Q = 6$ cm (tangents from $C$ ).
$D R = C D - C R = 9 - 6 = 3$ cm, and $D S = D R = 3$ cm (tangents from $D$ ).
$A D = A S + S D = 6 + 3 = 9$ cm.
Check: $A B + C D = 18 + 9 = 27$ and $A D + B C = 9 + 18 = 27$ . Correct.

Answer: $A B = 18$ cm and $A D = 9$ cm.

Maths in real life: Circles

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Secant, tangent and the point of contact

Imagine your colony park has a circular flower bed of radius 10 m, with its centre marked by a small fountain. The gardener plans three straight footpaths. Their perpendicular distances from the fountain are 12 m, 10 m and 6 m.

For each footpath, decide whether it misses the flower bed, just touches it or cuts through it. For the path that cuts through, find the length of that path lying inside the flower bed.

Try it first, then show the solutionSolution

Compare each perpendicular distance $d$ with the radius $r = 10$ m.
Path 1: $d = 12 > 10$ , so it is a non-intersecting line and misses the bed.
Path 2: $d = 10 = r$ , so it is a tangent and touches the bed at exactly one point.
Path 3: $d = 6 < 10$ , so it is a secant and cuts the bed at two points, say $A$ and $B$ .
Let $M$ be the foot of the perpendicular from the centre $O$ to $A B$ . The perpendicular from the centre to a chord bisects the chord, so $A M = M B$ .
In right $△ O M A$ : $A M^{2} = O A^{2} - O M^{2} = 1 0^{2} - 6^{2} = 100 - 36 = 64$ .
So $A M = 8$ m.
$A B = 2 \times A M = 2 \times 8 = 16$ m.

Answer: The 12 m path misses the bed, the 10 m path is a tangent and the 6 m path is a secant. The length of the 6 m path inside the bed is 16 m.

Memory hook: Fountain-to-path distance against radius: more misses, equal touches, less cuts.

Using the right angle: Pythagoras with tangents

Suppose a village pond is circular with radius 7 m. Ravi stands on the bank road at a point $P$ that is 18 m from the nearest point of the pond's edge. From $P$ he looks along a straight line that just grazes the edge of the pond at a point $T$ .

Find the distance from Ravi to the point $T$ where his line of sight touches the pond.

Try it first, then show the solutionSolution

Let $O$ be the centre of the pond. The nearest point $A$ of the edge lies on the line $P O$ , so $O P = P A + O A = 18 + 7 = 25$ m.
The line of sight $P T$ is a tangent at $T$ , so $O T ⊥ P T$ (tangent is perpendicular to the radius at the point of contact).
So $△ O T P$ is right-angled at $T$ , with hypotenuse $O P$ .
$P T^{2} = O P^{2} - O T^{2} = 2 5^{2} - 7^{2} = 625 - 49 = 576$ .
$P T = 576 = 24$ m.
Check: $7^{2} + 2 4^{2} = 49 + 576 = 625 = 2 5^{2}$ .

Answer: Ravi is 24 m from the point of contact $T$ .

Memory hook: Distance to the edge plus the radius gives the hypotenuse. Then Pythagoras gives the tangent.

Number of tangents from a point

Imagine a coach draws the school cricket ground on a grid. The boundary rope is a circle with centre $O (0, 0)$ and radius 5 units. A fielder stands at $F (13, 0)$ , and the coach asks him to sprint along straight lines that just touch the boundary rope.

How many such straight runs (tangents to the boundary) are possible from $F$ ? Find the length of each run from $F$ to the point where it touches the rope, and the area of the quadrilateral formed by the fielder, the two points of contact and the centre $O$ .

Try it first, then show the solutionSolution

By the distance formula, $O F = (13 - 0)^{2} + (0 - 0)^{2} = 13$ units.
$13 > 5$ , so $F$ lies outside the circle and exactly 2 tangents can be drawn. Let them touch the rope at $A$ and $B$ .
$O A ⊥ F A$ (tangent is perpendicular to the radius at the point of contact), so $△ O A F$ is right-angled at $A$ with hypotenuse $O F$ .
$F A^{2} = O F^{2} - O A^{2} = 1 3^{2} - 5^{2} = 169 - 25 = 144$ , so $F A = 12$ units.
$F B = F A = 12$ units (tangents from an external point are equal).
$O F$ splits quadrilateral $O A F B$ into $△ O A F$ and $△ O B F$ , which are congruent by RHS.
Area of $△ O A F = \frac{1}{2} \times O A \times F A = \frac{1}{2} \times 5 \times 12 = 30$ square units.
Area of $O A F B = 2 \times 30 = 60$ square units.
Check: $5^{2} + 1 2^{2} = 25 + 144 = 169 = 1 3^{2}$ . Correct.

Answer: 2 tangent runs are possible, each 12 units long, and the quadrilateral $O A F B$ has area 60 square units.

Memory hook: Outside the rope: two equal runs, and the kite they make is two right triangles.

Theorem 2: tangents from an external point are equal (proof)

Suppose two friends start together from a temple gate $P$ . Each walks along a different straight path, and each path just touches a circular pond of radius 5 m, one at $A$ and the other at $B$ . One friend counts his walk as $(2 x + 4)$ m and the other as $(4 x - 4)$ m.

Find $x$ and the length of each walk. Then find the distance of the gate from the centre of the pond and from the nearest point of its edge.

Try it first, then show the solutionSolution

$P A$ and $P B$ are tangents from the same external point $P$ , so $P A = P B$ (Theorem 2).
$2 x + 4 = 4 x - 4$ , so $2 x = 8$ and $x = 4$ .
$P A = 2 (4) + 4 = 12$ m and $P B = 4 (4) - 4 = 12$ m.
Let $O$ be the centre. Then $O A ⊥ P A$ (tangent is perpendicular to the radius), so $△ O A P$ is right-angled at $A$ .
$O P^{2} = O A^{2} + P A^{2} = 25 + 144 = 169$ , so $O P = 13$ m.
The nearest point of the edge lies on $O P$ , so its distance from $P$ is $O P - r = 13 - 5 = 8$ m.

Answer: $x = 4$ , and each walk is 12 m. The gate is 13 m from the centre and 8 m from the nearest point of the pond's edge.

Memory hook: Two friends grazing the same pond from one gate always walk the same distance.

Angle between two tangents and the angle at the centre

Imagine a town where two straight roads leave a junction $P$ . Each road just touches a circular roundabout, at $A$ and $B$ . The angle between the two roads at $P$ is $6 0^{\circ}$ , and each road segment $P A$ and $P B$ is 21 m long.

Find $∠ A O B$ at the centre $O$ of the roundabout, $∠ P A B$ , $∠ O A B$ , and the length of a straight cable laid from $A$ to $B$ .

Try it first, then show the solutionSolution

$∠ O A P = ∠ O B P = 9 0^{\circ}$ , because each tangent is perpendicular to the radius.
In quadrilateral $O A P B$ : $∠ A O B = 36 0^{\circ} - 9 0^{\circ} - 9 0^{\circ} - 6 0^{\circ} = 12 0^{\circ}$ .
$P A = P B$ (tangents from an external point), so $△ P A B$ is isosceles and $∠ P A B = ∠ P B A = \frac{18 0 ^{\circ} - 6 0 ^{\circ}}{2} = 6 0^{\circ}$ .
$∠ O A B = 9 0^{\circ} - 6 0^{\circ} = 3 0^{\circ}$ . Check: this is $\frac{1}{2} \times 6 0^{\circ}$ .
All three angles of $△ P A B$ are $6 0^{\circ}$ , so the triangle is equilateral.
$A B = P A = 21$ m.

Answer: $∠ A O B = 12 0^{\circ}$ , $∠ P A B = 6 0^{\circ}$ , $∠ O A B = 3 0^{\circ}$ , and the cable $A B$ is 21 m long.

Memory hook: Road angle plus centre angle make a straight $18 0^{\circ}$ .

Triangle drawn around a circle (incircle)

Suppose a housing society has a triangular garden $A B C$ with $A B = 13$ m, $B C = 14$ m and $C A = 15$ m. A circular fountain basin of radius 4 m is built so that it touches all three boundary walls: at $D$ on $B C$ , $E$ on $C A$ and $F$ on $A B$ .

Find the tangent lengths $A F$ , $B D$ and $C E$ along the walls from each corner to the points where the basin touches. Then find the area of the garden.

Try it first, then show the solutionSolution

Tangents from a vertex are equal, so $A F = A E = x$ , $B F = B D = y$ and $C D = C E = z$ .
$x + y = 13$ , $y + z = 14$ and $z + x = 15$ .
Adding all three: $2 (x + y + z) = 42$ , so $x + y + z = 21 = s$ .
$x = 21 - 14 = 7$ m, $y = 21 - 15 = 6$ m, $z = 21 - 13 = 8$ m.
Check: $x + y = 13$ , $y + z = 14$ , $z + x = 15$ , all correct.
Join the centre to $A$ , $B$ and $C$ . This splits the garden into three triangles, each of height $r = 4$ m.
Area $= \frac{1}{2} \times 4 \times (13 + 14 + 15) = 2 \times 42 = 84$ m².

Answer: $A F = 7$ m, $B D = 6$ m, $C E = 8$ m, and the area of the garden is 84 m².

Memory hook: From each corner to the basin: semi-perimeter minus the opposite wall.

Topics in this chapter

Circles formula sheet

Theorem 1: tangent perpendicular to radius: $O T ⊥ P T, ∠ O T P = 9 0^{\circ}$; Quote this as the reason every time you use a right angle at a point of contact.
Tangent length from an external point: $P T^{2} = O P^{2} - r^{2}$; From right triangle OTP. OP (centre to external point) is always the hypotenuse.
Theorem 2: equal tangents: $P A = P B$; Two tangents from the same external point are equal. Proof uses RHS congruence.
Centre line bisects the angles: $∠ A P O = ∠ B P O, ∠ A O P = ∠ B O P$; From the same congruence. Use when one angle between tangents is given and you need half of it.
OP is the perpendicular bisector of the chord of contact: $O P ⊥ A B, A M = M B$; M is where OP cuts AB. Useful for finding the length of AB.
Angle between tangents and angle at centre: $∠ A P B + ∠ A O B = 18 0^{\circ}$; From quadrilateral OAPB with two right angles. The commonest MCQ fact.
Angles with the chord of contact: $∠ P A B = ∠ P B A = 9 0^{\circ} - \frac{1}{2} ∠ A P B, ∠ O A B = \frac{1}{2} ∠ A P B$; Derive in the exam from isosceles triangles PAB and OAB; quote only in MCQs.
Number of tangents from a point: $inside: 0, on: 1, outside: 2$; Decide position by comparing distance from centre with the radius.
Tangents at ends of a diameter: $l ∥ m, distance = 2 r$; Also: if two parallel lines both touch a circle, the distance between them is the diameter.
Chord of larger circle touching smaller concentric circle: $A B = 2 R^{2} - r^{2}$; The chord is bisected at the point of contact. R is the larger radius.
Quadrilateral circumscribing a circle: $A B + C D = A D + B C$; Sums of opposite sides are equal. A circumscribed parallelogram is a rhombus.
Tangent lengths in a triangle with an incircle: $A F = s - B C, B D = s - C A, C E = s - A B, s = \frac{A B + B C + C A}{2}$; Each vertex's tangent length is the semi-perimeter minus the opposite side.
Inradius of a right triangle: $r = \frac{a + b - c}{2}$; a and b are the legs, c the hypotenuse. Comes from the square formed at the right angle.
Area of a triangle using its incircle: $Area = \frac{1}{2} r (a + b + c)$; Join the centre to the vertices; three triangles each have height r.
Perimeter of triangle cut off by a third tangent: $P C + C D + D P = 2 P A$; PA, PB tangents; CD touches the circle between them, with C on PA and D on PB.

How the board asks Circles

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Angle chasing with two tangents	An MCQ or 2-mark question gives one angle, such as the angle between the tangents, the angle at the centre, or an angle between a chord and a radius, and asks for another.	Join the centre to both points of contact. Mark the two $9 0^{\circ}$ angles. Use $∠ A P B + ∠ A O B = 18 0^{\circ}$ and the isosceles triangles $P A B$ and $O A B$ . Check your answer with $∠ O A B = \frac{1}{2} ∠ A P B$ .
Finding tangent length, radius or distance from centre	Two of tangent length, radius and distance $O P$ are given directly or in disguise (such as the distance from $P$ to the nearest point of the circle), and the third is asked. Also concentric circle chord lengths.	Draw right $△ O T P$ with the right angle at $T$ and $O P$ as hypotenuse. Write the reason for the right angle, then apply Pythagoras. If a distance to the circle is given, add the radius to get $O P$ .
Proving the two theorems	A 3- or 5-mark question asks you to prove that the tangent is perpendicular to the radius, or that tangents from an external point are equal, often followed by a short application.	Write Given, To prove, Construction and Proof with a described figure. For Theorem 1 use the shortest-distance argument. For Theorem 2 use RHS congruence with three named pairs and finish with CPCT.
Proofs built on the theorems	Prove that tangents at ends of a diameter are parallel, that a circumscribed parallelogram is a rhombus, that $∠ A P B = 2∠ O A B$ , or that opposite sides of a circumscribed quadrilateral subtend supplementary angles at the centre.	Start by listing every pair of equal tangents or every right angle the figure gives you. Then add equations or angles. Give a reason in brackets on every line.
Triangles and quadrilaterals around a circle	Some side lengths or tangent segments are given and you find a missing side, a tangent segment, the perimeter or the inradius.	Label equal tangents from each vertex with the same letter. Write one equation per side and solve. For quadrilaterals use $A B + C D = A D + B C$ ; for right angles at a vertex, remember the tangent segments there equal $r$ .
Assertion-reason on tangent properties	The assertion is a numerical or geometric claim (for example about an angle or the number of tangents) and the reason is a property of tangents.	Check A independently by working it out. Check R as a general fact. Then ask whether R is the actual property you used to get A.
Perimeter of the triangle cut off by a third tangent	Tangents $P A$ and $P B$ are given; a third tangent touches the circle between them and meets $P A$ and $P B$ , and you find the perimeter of the small triangle.	Split the third tangent at its point of contact. Replace each piece by its equal tangent on $P A$ or $P B$ . The perimeter becomes $P A + P B = 2 P A$ .
Case study from real life	A Section E context such as a wheel on a road, a belt around a pulley, a ring touching the sides of a frame, or a circular track with straight paths, with sub-parts of 1, 1 and 2 marks.	Translate the picture into a centre, radius and tangent lines. Identify every right angle and pair of equal tangents first, then answer each sub-part with the matching property.

Tricks to remember Circles

Short memory aids. Say each one aloud once, then use it on a question the same day.

Radius meets tangent square-on
Whenever you see a point of contact, draw the radius to it and put a right-angle mark. No radius, no right angle, no marks.
Use it when: Your first move on every figure; it unlocks Pythagoras and angle sums.
The kite OAPB
Centre, two points of contact and the external point form a kite with two right angles. Right angles add to $18 0^{\circ}$ , so the other two angles add to $18 0^{\circ}$ too.
Use it when: Instantly gives $∠ A P B + ∠ A O B = 18 0^{\circ}$ in MCQs.
Half the outside angle
The angle between the chord of contact and a radius is half the angle between the tangents: $∠ O A B = \frac{1}{2} ∠ A P B$ .
Use it when: A quick check on any angle-chasing answer.
Opposite sides add up
For a quadrilateral around a circle, cross the sides: top plus bottom equals left plus right. Adjacent sides never work.
Use it when: Finding a missing side, or proving a parallelogram is a rhombus.
s minus the opposite side
In a triangle with an incircle, the tangent length from a vertex is the semi-perimeter minus the side facing that vertex.
Use it when: Gets all three tangent lengths in one line, then check they add to each side.
Hypotenuse goes to the centre
In the tangent triangle, the longest side always runs from the external point to the centre. So $O P^{2} = r^{2} + P T^{2}$ , never $P T^{2} = O P^{2} + r^{2}$ .
Use it when: Stops the most common Pythagoras slip.

Circles: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Angle at the centre in a given ratio · 1 mark · MCQBoard level

$P A$ and $P B$ are tangents from an external point $P$ to a circle with centre $O$ , touching it at $A$ and $B$ . If $∠ A O B : ∠ A P B = 7 : 2$ , then $∠ A P B$ is
(a) $2 0^{\circ}$
(b) $4 0^{\circ}$
(c) $14 0^{\circ}$
(d) $8 0^{\circ}$

Try it first, then show the solutionSolution

$∠ O A P = ∠ O B P = 9 0^{\circ}$ (tangent is perpendicular to the radius at the point of contact).
Angles of quadrilateral $O A P B$ add to $36 0^{\circ}$ , so $∠ A O B + ∠ A P B = 18 0^{\circ}$ .
Let $∠ A O B = 7 k$ and $∠ A P B = 2 k$ . Then $9 k = 18 0^{\circ}$ , so $k = 2 0^{\circ}$ .
$∠ A P B = 2 k = 4 0^{\circ}$ .
Check: $∠ A O B = 14 0^{\circ}$ and $14 0^{\circ} + 4 0^{\circ} = 18 0^{\circ}$ . Option (c) is the trap: it is $∠ A O B$ , not $∠ A P B$ .

Answer: (b) $4 0^{\circ}$

Example 2: Isosceles triangle formed by two tangents · 1 mark · Assertion-reasonBoard level

$P A$ and $P B$ are tangents from an external point $P$ to a circle, touching it at $A$ and $B$ .
Assertion (A): If $∠ A P B = 5 0^{\circ}$ , then $∠ P A B = 6 5^{\circ}$ .
Reason (R): $P A = P B$ , so $△ P A B$ is isosceles.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Try it first, then show the solutionSolution

R: tangents from an external point are equal, so $P A = P B$ and $△ P A B$ is isosceles. R is true.
Since $P A = P B$ , $∠ P A B = ∠ P B A$ (angles opposite equal sides).
$∠ P A B = \frac{18 0 ^{\circ} - 5 0 ^{\circ}}{2} = 6 5^{\circ}$ . A is true.
The value in A comes directly from the isosceles triangle in R, so R explains A.

Answer: (a) Both A and R are true and R is the correct explanation of A.

Example 3: Perimeter of the triangle cut off by a third tangent · 3 marks · Short answerHarder board level

A circle has centre $O$ and radius $8$ cm. A point $P$ is $17$ cm from $O$ . $P A$ and $P B$ are the tangents from $P$ , touching the circle at $A$ and $B$ . A third tangent touches the circle at a point $E$ on the minor arc $A B$ (the arc nearer to $P$ ) and meets $P A$ at $C$ and $P B$ at $D$ . Find the perimeter of $△ P C D$ . If $C E = 4$ cm, also find $P C$ .

Try it first, then show the solutionSolution

$O A ⊥ P A$ (tangent is perpendicular to the radius), so in right $△ O A P$ : $P A = 1 7^{2} - 8^{2} = 289 - 64 = 225 = 15$ cm. Also $P B = P A = 15$ cm (equal tangents from $P$ ). [1 mark]
From $C$ : $C A = C E$ ; from $D$ : $D B = D E$ (tangents from an external point are equal). [½ mark]
Perimeter of $△ P C D = P C + C E + E D + D P = P C + C A + D B + D P$ .
$= (P C + C A) + (D B + D P) = P A + P B = 15 + 15 = 30$ cm. [1 mark]
$C A = C E = 4$ cm, so $P C = P A - C A = 15 - 4 = 11$ cm. [½ mark]
Check: with $P C = 11$ and $C E = 4$ , $P C + C E = 15 = P A$ . Correct.

Answer: Perimeter of $△ P C D = 30$ cm and $P C = 11$ cm.

Example 4: Tangent at D bisects BC · 3 marks · Short answerHarder board level

$△ A B C$ is right-angled at $B$ . A circle drawn with $A B$ as diameter meets the hypotenuse $A C$ at $D$ . The tangent to the circle at $D$ meets $B C$ at $E$ . Prove that $E$ is the midpoint of $B C$ .

Try it first, then show the solutionSolution

The centre $O$ is the midpoint of $A B$ , so $O B$ is a radius. Since $∠ A B C = 9 0^{\circ}$ , $B C ⊥ O B$ at $B$ , so $B C$ is the tangent at $B$ (a line through the end of a radius and perpendicular to it is a tangent).
$E B$ and $E D$ are tangents from the external point $E$ , so $E B = E D$ . [1 mark]
$∠ A D B = 9 0^{\circ}$ (angle in a semicircle), so $∠ B D C = 9 0^{\circ}$ . [½ mark]
In $△ E B D$ , $E B = E D$ , so $∠ E B D = ∠ E D B$ .
In right $△ B D C$ : $∠ E C D = 9 0^{\circ} - ∠ E B D$ . Also $∠ E D C = 9 0^{\circ} - ∠ E D B$ (as $∠ B D C = 9 0^{\circ}$ ).
So $∠ E C D = ∠ E D C$ , hence $E D = E C$ (sides opposite equal angles). [1 mark]
Therefore $E B = E D = E C$ , so $E$ is the midpoint of $B C$ . Hence proved. [½ mark]

Answer: $E B = E D$ (equal tangents) and $E D = E C$ (equal base angles), so $E B = E C$ and the tangent at $D$ bisects $B C$ .

Example 5: Proof of Theorem 1 with a quadratic application · 5 marks · Long answerHarder board level

(i) Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
(ii) $P T$ is a tangent from a point $P$ to a circle with centre $O$ , touching it at $T$ . $O P = 13$ cm and the length of the tangent $P T$ is $7$ cm more than the radius. Find the radius and $P T$ .

Try it first, then show the solutionSolution

Given: a circle with centre $O$ and a tangent $X Y$ touching it at $P$ . To prove: $O P ⊥ X Y$ . Figure: draw the circle, the tangent $X Y$ at $P$ , and join $O P$ . [1 mark]
Construction: take any point $Q$ on $X Y$ other than $P$ and join $O Q$ .
A tangent meets the circle at exactly one point, so $Q$ lies outside the circle. Hence $O Q$ is longer than the radius: $O Q > O P$ . [1 mark]
This is true for every point $Q$ on $X Y$ except $P$ . So $O P$ is the shortest of all segments from $O$ to points of $X Y$ .
The shortest segment from a point to a line is the perpendicular. So $O P ⊥ X Y$ . Hence proved. [1 mark]
(ii) Let the radius $O T = r$ cm, so $P T = (r + 7)$ cm. $O T ⊥ P T$ by part (i), so $O T^{2} + P T^{2} = O P^{2}$ . [½ mark]
$r^{2} + (r + 7)^{2} = 169 \Rightarrow 2 r^{2} + 14 r + 49 - 169 = 0 \Rightarrow r^{2} + 7 r - 60 = 0$ .
$(r + 12) (r - 5) = 0$ , so $r = 5$ or $r = - 12$ . A radius cannot be negative, so $r = 5$ cm. [1 mark]
$P T = 5 + 7 = 12$ cm. Check: $5^{2} + 1 2^{2} = 25 + 144 = 169 = 1 3^{2}$ . [½ mark]

Answer: (i) Proved: $O P$ is the shortest distance from $O$ to the tangent, so it is perpendicular to it. (ii) Radius $= 5$ cm and $P T = 12$ cm.

Example 6: Sides of a right triangle from its incircle · 5 marks · Long answerHarder board level

A circle with centre $O$ and radius $4$ cm is inscribed in $△ A B C$ , right-angled at $B$ . It touches $B C$ , $C A$ and $A B$ at $D$ , $E$ and $F$ respectively. If $C D = 6$ cm, find the lengths $A B$ , $B C$ and $A C$ .

Try it first, then show the solutionSolution

$O D ⊥ B C$ and $O F ⊥ A B$ (radius is perpendicular to the tangent at the point of contact). With $∠ B = 9 0^{\circ}$ , quadrilateral $O F B D$ has all four angles $9 0^{\circ}$ . [1 mark]
Also $O D = O F = 4$ cm (radii), so $O F B D$ is a square and $B D = B F = 4$ cm. [1 mark]
Tangents from an external point are equal: $C E = C D = 6$ cm, and let $A E = A F = x$ cm. [½ mark]
So $B C = B D + D C = 4 + 6 = 10$ cm, $A B = x + 4$ and $A C = x + 6$ . [½ mark]
By Pythagoras in $△ A B C$ : $(x + 6)^{2} = (x + 4)^{2} + 1 0^{2}$ . [½ mark]
$x^{2} + 12 x + 36 = x^{2} + 8 x + 16 + 100 \Rightarrow 4 x = 80 \Rightarrow x = 20$ . [1 mark]
$A B = 24$ cm, $B C = 10$ cm, $A C = 26$ cm.
Check: $2 4^{2} + 1 0^{2} = 576 + 100 = 676 = 2 6^{2}$ , and $r = \frac{24 + 10 - 26}{2} = 4$ cm. Both correct. [½ mark]

Answer: $A B = 24$ cm, $B C = 10$ cm, $A C = 26$ cm

Example 7: Find the error: circumscribed quadrilateral · 3 marks · Short answerHOTS / competency

Quadrilateral $A B C D$ is drawn to circumscribe a circle with centre $O$ , touching $A B$ , $B C$ , $C D$ , $D A$ at $P$ , $Q$ , $R$ , $S$ respectively. $A B = 10$ cm, $B C = 13$ cm, $C D = 15$ cm, $∠ A = 9 0^{\circ}$ and $A P = 6$ cm. A student writes: " $A B + B C = C D + D A$ , so $23 = 15 + D A$ , giving $D A = 8$ cm."
(i) Find the error and give the correct $D A$ and the perimeter of $A B C D$ .
(ii) Find $B P$ , $C Q$ and the radius of the circle.

Try it first, then show the solutionSolution

Equal tangents from each vertex give $A P = A S$ , $B P = B Q$ , $C R = C Q$ , $D R = D S$ .
Adding: $A B + C D = (A P + B P) + (C R + D R) = (A S + B Q) + (C Q + D S) = A D + B C$ . The equal sums are of opposite sides; the student added adjacent sides.
Correct: $10 + 15 = D A + 13 \Rightarrow D A = 12$ cm, and perimeter $= 10 + 13 + 15 + 12 = 50$ cm. [1 mark]
(ii) $O P ⊥ A B$ and $O S ⊥ A D$ (tangent is perpendicular to radius), and $∠ A = 9 0^{\circ}$ , so $A P O S$ has four right angles with $O P = O S$ (radii). It is a square, so $r = A P = 6$ cm. [1 mark]
$B P = A B - A P = 10 - 6 = 4$ cm, so $B Q = B P = 4$ cm (tangents from $B$ ).
$C Q = B C - B Q = 13 - 4 = 9$ cm. [1 mark]
Check: $A S = 6$ , so $D S = 12 - 6 = 6 = D R$ , and $C R = 15 - 6 = 9 = C Q$ . All consistent.

Answer: (i) The error is pairing adjacent sides; the rule is $A B + C D = A D + B C$ , so $D A = 12$ cm and perimeter $= 50$ cm. (ii) $B P = 4$ cm, $C Q = 9$ cm and radius $= 6$ cm.

Example 8: A rectangle around a circle must be a square · 3 marks · Short answerHOTS / competency

(i) Prove that a rectangle that circumscribes a circle must be a square.
(ii) A square wooden frame is made to fit exactly around a circular mirror of radius $21$ cm, each side touching the mirror. Find the perimeter of the inner edge of the frame.

Try it first, then show the solutionSolution

(i) Let rectangle $A B C D$ touch the circle at $P$ , $Q$ , $R$ , $S$ on $A B$ , $B C$ , $C D$ , $D A$ . Equal tangents from $A$ , $B$ , $C$ , $D$ give $A B + C D = A D + B C$ . [1 mark]
In a rectangle $A B = C D$ and $A D = B C$ , so $2 A B = 2 B C$ , that is $A B = B C$ .
A rectangle with two adjacent sides equal is a square. Hence proved. [1 mark]
(ii) Opposite sides of the square are parallel tangents, so the distance between them is the diameter: side $= 2 \times 21 = 42$ cm.
Perimeter $= 4 \times 42 = 168$ cm. [1 mark]

Answer: (i) Proved: $A B + C D = A D + B C$ with opposite sides equal forces $A B = B C$ . (ii) Perimeter $= 168$ cm.

Case-study practice (Section E style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 1 + 2 marks.

Pipes around a circular water tank

A village in Rajasthan has a circular water tank. Seen from above, its rim is a circle with centre $O$ and radius $7$ m. A pump house stands at point $P$ , $25$ m from $O$ . Two straight pipes run from $P$ and just touch the rim, one at $A$ and one at $B$ , so $P A$ and $P B$ are tangents to the circle. The panchayat engineer needs the pipe lengths, the gap between the pump house and the tank, and the size of the land enclosed by the two pipes and the two radii $O A$ and $O B$ .

(i) · 1 mark
Find the length of each pipe.
Answer
Answer: $P A = P B = 24$ m
$O A ⊥ P A$ (tangent is perpendicular to the radius), so $P A = 2 5^{2} - 7^{2} = 625 - 49 = 576 = 24$ m. $P B = P A$ as tangents from $P$ are equal.
(ii) · 1 mark
Find the shortest distance from the pump house $P$ to the rim of the tank.
Answer
Answer: $18$ m
The nearest point of the rim lies on $P O$ , so the distance is $O P - r = 25 - 7 = 18$ m.
(iii) · 2 marks
Find the area of quadrilateral $P A O B$ and hence the length of the chord $A B$ .
Answer
Answer: Area $= 168$ m² and $A B = 13.44$ m
Area $= 2 \times \frac{1}{2} \times O A \times P A = 7 \times 24 = 168$ m². Since $P A = P B$ and $O A = O B$ , $O P$ is the perpendicular bisector of $A B$ , so area $= \frac{1}{2} \times O P \times A B$ . Then $168 = \frac{1}{2} \times 25 \times A B$ , giving $A B = \frac{336}{25} = 13.44$ m.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Two concentric circles have radii $5$ cm and $3$ cm. Find the length of the chord of the larger circle which touches the smaller circle. (2 marks)

Step	Marks
Let the chord $A B$ of the larger circle touch the smaller circle at $P$ . $O P ⊥ A B$ (tangent is perpendicular to radius), and the perpendicular from the centre bisects the chord, so $A P = P B$ .	0.5
In right $△ O P A$ : $A P^{2} = O A^{2} - O P^{2} = 25 - 9 = 16$ , so $A P = 4$ cm.	1
$A B = 2 A P = 8$ cm.	0.5

A circle is inscribed in $△ A B C$ , right-angled at $B$ , with $A B = 24$ cm and $B C = 7$ cm. Find the radius of the circle. (3 marks)

Step	Marks
$A C = 2 4^{2} + 7^{2} = 625 = 25$ cm. Let the circle (centre $O$ , radius $r$ ) touch $A B$ , $B C$ , $C A$ at $P$ , $Q$ , $R$ . $O P ⊥ A B$ , $O Q ⊥ B C$ and $∠ B = 9 0^{\circ}$ , with $O P = O Q$ , so $O P B Q$ is a square and $B P = B Q = r$ .	1
$A P = 24 - r$ and $C Q = 7 - r$ . Tangents from an external point are equal, so $A R = A P = 24 - r$ and $C R = C Q = 7 - r$ .	1
$A C = A R + R C$ : $25 = 31 - 2 r$ , so $r = 3$ cm.	1

Prove that the lengths of tangents drawn from an external point to a circle are equal. Using this, prove that if a quadrilateral $A B C D$ is drawn to circumscribe a circle, then $A B + C D = A D + B C$ . (5 marks)

Step	Marks
Given: circle with centre $O$ , tangents $P A$ and $P B$ from external point $P$ touching at $A$ and $B$ . To prove: $P A = P B$ . Figure described.	1
Construction: join $O A$ , $O B$ and $O P$ .	0.5
$∠ O A P = ∠ O B P = 9 0^{\circ}$ (tangent is perpendicular to radius through point of contact).	1
In $△ O A P$ and $△ O B P$ : $O A = O B$ (radii), $O P = O P$ (common), so $△ O A P ≅ △ O B P$ (RHS).	1
Hence $P A = P B$ (CPCT).	0.5
For $A B C D$ touching the circle at $P, Q, R, S$ on $A B, B C, C D, D A$ : $A P = A S$ , $B P = B Q$ , $C R = C Q$ , $D R = D S$ . Adding: $(A P + B P) + (C R + D R) = (A S + D S) + (B Q + C Q)$ , so $A B + C D = A D + B C$ .	1

Where students lose marks in Circles

The slips examiners see most often, and what to write instead.

Mistake: Treating the angle at the centre as equal to, or double, the angle between the tangents
Students mix this up with the inscribed-angle rule from Class 9, where the angle at the centre is twice the angle on the circle.
Wrong working
$∠ A P B = 7 0^{\circ}$ , so $∠ A O B = 2 \times 7 0^{\circ} = 14 0^{\circ}$
Right working
$∠ O A P = ∠ O B P = 9 0^{\circ}$ , so $∠ A O B = 36 0^{\circ} - 9 0^{\circ} - 9 0^{\circ} - 7 0^{\circ} = 11 0^{\circ}$
Do this: Always use the quadrilateral $O A P B$ : the two angles are supplementary.
Usually costs: The full 1 mark in an MCQ, and all later steps in a longer question.
Mistake: Putting the tangent as the hypotenuse
The tangent is drawn as the longest-looking line, so students add instead of subtract.
Wrong working
$P T^{2} = O P^{2} + O T^{2} = 169 + 25 = 194$
Right working
$∠ O T P = 9 0^{\circ}$ , so $O P$ is the hypotenuse: $P T^{2} = O P^{2} - O T^{2} = 169 - 25 = 144$ , $P T = 12$ cm
Do this: Mark the right angle at $T$ first. The side opposite it, $O P$ , is the hypotenuse.
Usually costs: 1 to 2 marks; only the reason step survives.
Mistake: Taking the distance from the external point to the circle as the distance to the centre
The question says the point is a certain distance 'from the circle', and students read it as $O P$ .
Wrong working
$P A = 8$ , so $O P = 8$ and $P T^{2} = 64 - r^{2}$
Right working
$O P = P A + A O = 8 + r$ , so $(8 + r)^{2} = r^{2} + P T^{2}$
Do this: Draw line $P O$ through the circle and mark where it meets the circle before writing any equation.
Usually costs: Most of a 3-mark question.
Mistake: Missing the reasons in the proof of Theorem 1
Students write the conclusion without saying why $O Q$ is longer than $O P$ .
Wrong working
$O Q > O P$ . So $O P ⊥ X Y$ .
Right working
$Q$ lies outside the circle since the tangent meets the circle only at $P$ , so $O Q > O P$ . This holds for every point $Q$ on $X Y$ except $P$ , so $O P$ is the shortest distance from $O$ to $X Y$ , and the shortest segment from a point to a line is perpendicular to it. Hence $O P ⊥ X Y$ .
Do this: Learn the proof as a chain of three reasons, not as two lines.
Usually costs: 1 to 2 marks in a 3- or 5-mark proof.
Mistake: Wrong or unsupported congruence in the proof of Theorem 2
Students write SAS or SSS, or write RHS without saying why the angles are right angles.
Wrong working
$O A = O B$ , $O P = O P$ , so $△ O A P ≅ △ O B P$ (SAS)
Right working
$∠ O A P = ∠ O B P = 9 0^{\circ}$ (tangent $⊥$ radius), $O A = O B$ (radii), $O P = O P$ (common hypotenuse), so $△ O A P ≅ △ O B P$ (RHS); hence $P A = P B$ (CPCT)
Do this: Name all three pairs with a reason each, write RHS, then CPCT.
Usually costs: 1 to 2 marks.
Mistake: Adding adjacent sides of a circumscribed quadrilateral
The equal-tangent equations are added in a muddle and grouped wrongly.
Wrong working
$A B + B C = C D + D A$
Right working
$A P + B P + C R + D R = A S + B Q + C Q + D S$ , so $A B + C D = A D + B C$
Do this: Write the four equal-tangent pairs in a column, add them, then group into whole sides.
Usually costs: 1 to 2 marks, and a wrong final answer.

NCERT exercise guide: Circles

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Exercise 10.1

What it tests: The basic language of the chapter: what a tangent and a secant are, how many tangents a circle can have, how many tangents a circle has at one point, and how many lines parallel to a given line can touch a circle. It also asks for a tangent length using the right angle between the radius and the tangent.

How to approach it: Draw a rough figure every time, even for a one-word answer. For tangent lengths, join the centre to the point of contact, mark the right angle there, and apply Pythagoras with the line from the centre to the external point as the hypotenuse. Always name the reason for the right angle: the tangent is perpendicular to the radius at the point of contact.

Like the tangent-length question in Exercise 10.1

A tangent $P Q$ at a point $P$ of a circle of radius $9$ cm meets a line through the centre $O$ at a point $Q$ such that $O Q = 15$ cm. Find the length of $P Q$ .

Show the solutionSolution

Join $O P$ . $O P = 9$ cm is the radius through the point of contact $P$ .
$O P ⊥ P Q$ (the tangent at any point of a circle is perpendicular to the radius through the point of contact), so $∠ O P Q = 9 0^{\circ}$ .
In right $△ O P Q$ , by Pythagoras: $P Q^{2} = O Q^{2} - O P^{2}$ .
$P Q^{2} = 1 5^{2} - 9^{2} = 225 - 81 = 144$ .
$P Q = 12$ cm.
Check: $9^{2} + 1 2^{2} = 81 + 144 = 225 = 1 5^{2}$ . Correct.

Answer: $P Q = 12$ cm

Exercise 10.2

What it tests: Using both theorems: tangent lengths and radii from given distances, angle chasing with two tangents from one point, proofs such as the tangents at the ends of a diameter being parallel, the angle between two tangents being twice an angle at the point of contact, and results for quadrilaterals and triangles drawn around a circle.

How to approach it: From any external point, mark the two equal tangent lengths on your figure first, then mark the right angles at the points of contact. For angle questions, use the quadrilateral formed by the centre, the external point and the two points of contact (its angles add to $36 0^{\circ}$ ) and the isosceles triangle formed by the two tangents and the chord. For figures drawn around a circle, write each side as a sum of two tangent lengths and add or compare.

Like the angle-chasing questions with two tangents in Exercise 10.2

$T P$ and $T Q$ are tangents from an external point $T$ to a circle with centre $O$ , touching it at $P$ and $Q$ . If $∠ P T Q = 6 4^{\circ}$ , find $∠ P O Q$ , $∠ T P Q$ and $∠ O P Q$ .

Show the solutionSolution

$O P ⊥ T P$ and $O Q ⊥ T Q$ (tangent is perpendicular to the radius at the point of contact), so $∠ O P T = ∠ O QT = 9 0^{\circ}$ .
In quadrilateral $O P T Q$ : $∠ P O Q = 36 0^{\circ} - 9 0^{\circ} - 9 0^{\circ} - 6 4^{\circ} = 11 6^{\circ}$ .
$T P = T Q$ (tangents from an external point are equal), so $△ T P Q$ is isosceles and $∠ T P Q = ∠ T QP$ .
$∠ T P Q = \frac{18 0 ^{\circ} - 6 4 ^{\circ}}{2} = 5 8^{\circ}$ .
$∠ O P Q = ∠ O P T - ∠ T P Q = 9 0^{\circ} - 5 8^{\circ} = 3 2^{\circ}$ .
Check: $∠ O P Q = 3 2^{\circ} = \frac{1}{2} \times 6 4^{\circ}$ , which matches the result $∠ P T Q = 2∠ O P Q$ .

Answer: $∠ P O Q = 11 6^{\circ}$ , $∠ T P Q = 5 8^{\circ}$ , $∠ O P Q = 3 2^{\circ}$

Circles: previous year questions

Standard: Asked in 21 of the 21 board papers we hold (2023 to 2026), on average 7.2 marks per paper; 90 questions in all, counting OR alternatives. In 4 of 4 sample papers.
Basic: Asked in 21 of the 21 board papers we hold (2023 to 2026), on average 6.5 marks per paper; 83 questions in all, counting OR alternatives. In 4 of 4 sample papers.

Counted from the real papers on our previous year papers page, not estimated. Average marks are over every board paper we hold, with an OR pair counted once. Chapter tags are ours.

Solved board questions from Circles

Real questions from recent papers. Try each one first, then open our step-by-step solution with marks per step.

2026 Basic, set 430/1/1, Q9 · 1 mark · MCQ

In the given figure, PA is a tangent to a circle with centre O. If $O P = 10$ cm, then the length of AP is :

(A) $103$ cm
(B) $20$ cm
(C) $5$ cm
(D) $53$ cm

Figure in the paper: Circle with centre O; A is a point of contact on the circle; P is outside the circle; PA is the tangent at A; OA is joined; OP = 10 cm; angle APO = 30° marked at P. See it in the official paper.

Try it first, then show our solutionSolution

$O A ⊥ P A$ [tangent ⟂ radius], so $△ O A P$ is right-angled at A.
½ mark
$A P = O P cos 3 0^{\circ} = 10 \times \frac{3}{2} = 53$ cm.
½ mark

Answer: 5√3 cm, option (D)

Where the marks are

Uses the right angle at A (½ mark)
Gets $A P = 53$ cm (½ mark)

See it in the full 2026 board paper, set 430/1/1

2026 Basic, set 430/1/1, Q23 · 2 marks · Very short answer

Two concentric circles are of radii 6 cm and 10 cm. Find the length of the chord of the larger circle which touches the smaller circle.

Try it first, then show our solutionSolution

Figure: two concentric circles with centre O. Chord AB of the larger circle touches the smaller circle at C. Join OC and OB. Here $O C = 6$ cm and $O B = 10$ cm.
½ mark
$O C ⊥ A B$ [tangent ⟂ radius at the point of contact]. In right $△ O C B$ : $B C^{2} = O B^{2} - O C^{2} = 1 0^{2} - 6^{2} = 64$ , so $B C = 8$ cm.
1 mark
The perpendicular from the centre bisects the chord, so $A B = 2 \times B C = 16$ cm.
½ mark

Answer: 16 cm

Where the marks are

Draws a correct figure (½ mark)
Uses $O C ⊥ A B$ and Pythagoras to get $B C = 8$ cm (1 mark)
Gets $A B = 16$ cm (½ mark)

See it in the full 2026 board paper, set 430/1/1

2026 Basic, set 430/1/1, Q29 · 3 marks · Short answer

Prove that a rectangle circumscribing a circle is a square.

Try it first, then show our solutionSolution

Figure: rectangle ABCD circumscribing a circle with centre O. The sides AB, BC, CD, DA touch the circle at P, Q, R, S respectively.
½ mark
Tangents drawn from an external point to a circle are equal, so $A P = A S$ , $B P = B Q$ , $C R = C Q$ and $D R = D S$ . Adding: $(A P + B P) + (C R + D R) = (A S + D S) + (B Q + C Q)$ , i.e. $A B + C D = A D + B C$ .
1 mark
In a rectangle, opposite sides are equal: $C D = A B$ and $B C = A D$ . So $2 A B = 2 A D \Rightarrow A B = A D$ .
1 mark
Adjacent sides of rectangle ABCD are equal, so ABCD is a square.
½ mark

Answer: AB + CD = AD + BC with AB = CD and AD = BC gives AB = AD, so ABCD is a square

Where the marks are

Draws a correct figure (½ mark)
Uses equal tangents to get $A B + C D = A D + B C$ (1 mark)
Uses opposite sides of the rectangle to show $A B = A D$ (1 mark)
Concludes ABCD is a square (½ mark)

See it in the full 2026 board paper, set 430/1/1

Every Circles question, by year

Standard (105 questions, sample papers included)

For 2027

Sample paper 2026-27: Q29 (3 marks, solved), Q37 (4 marks, solved)

2026

Set 30/1/1: Q12 (1 mark, solved), Q13 (1 mark, solved), Q25 (2 marks, solved), Q29(A) (3 marks, solved), Q29(B) (3 marks, solved)

Set 30/2/1: Q13 (1 mark, solved), Q14 (1 mark, solved), Q25 (2 marks, solved), Q29(a) (3 marks, solved), Q29(b) (3 marks, solved)

Set 30/3/1: Q2 (1 mark, solved), Q12 (1 mark, solved), Q19 (1 mark, solved), Q31(a) (3 marks, solved), Q31(b) (3 marks, solved)

Set 30/4/1: Q4 (1 mark, solved), Q14 (1 mark, solved), Q33 (5 marks, solved)

Set 30/5/1: Q4 (1 mark, solved), Q35 (5 marks, solved)

Sample paper 2025-26: Q4 (1 mark, solved), Q18 (1 mark, solved), Q25 (2 marks, solved), Q26 (3 marks, solved)

2025

Set 30/1/1: Q7 (1 mark, solved), Q16 (1 mark, solved), Q25 (2 marks, solved), Q26(a) (3 marks, solved), Q26(b) (3 marks, solved)

Set 30/2/1: Q7 (1 mark, solved), Q25 (2 marks, solved), Q26(a) (3 marks, solved), Q26(b) (3 marks, solved)

Set 30/3/1: Q8 (1 mark, solved), Q20 (1 mark, solved), Q25 (2 marks, solved), Q31 (3 marks, solved)

Set 30/4/1: Q14 (1 mark, solved), Q15 (1 mark, solved), Q20 (1 mark, solved), Q31 (3 marks, solved)

Set 30/5/1: Q10 (1 mark, solved), Q11 (1 mark, solved), Q14 (1 mark, solved), Q31 (3 marks, solved)

Set 30/6/1: Q9 (1 mark, solved), Q10 (1 mark, solved), Q11 (1 mark, solved), Q31 (3 marks, solved), Q32(a) (5 marks, solved)

Sample paper 2024-25: Q3 (1 mark, solved), Q7 (1 mark, solved), Q10 (1 mark, solved), Q33 (5 marks, solved)

2024

Set 30/1/1: Q19 (1 mark, solved), Q30(A) (3 marks, solved), Q30(B) (3 marks, solved), Q38 (4 marks, solved)

Set 30/2/1: Q14 (1 mark, solved), Q17 (1 mark, solved), Q24 (2 marks, solved), Q28(a) (3 marks, solved), Q28(b) (3 marks, solved)

Set 30/3/1: Q9 (1 mark, solved), Q10 (1 mark, solved), Q18 (1 mark, solved), Q31 (3 marks, solved)

Set 30/4/1: Q15 (1 mark, solved), Q18 (1 mark, solved), Q24 (2 marks, solved), Q31 (3 marks, solved)

Set 30/5/1: Q12 (1 mark, solved), Q16 (1 mark, solved), Q21(a) (2 marks, solved), Q21(b) (2 marks, solved), Q28 (3 marks, solved)

Sample paper 2023-24: Q9 (1 mark, solved), Q10 (1 mark, solved), Q23 (2 marks, solved), Q29(a) (3 marks, solved), Q29(b) (3 marks, solved)

2023

Set 30/1/1: Q5 (1 mark, solved), Q29(a) (3 marks, solved), Q29(b) (3 marks, solved), Q38 (4 marks, solved)

Set 30/2/1: Q15 (1 mark, solved), Q19 (1 mark, solved), Q25 (2 marks, solved), Q31 (3 marks, solved)

Set 30/4/1: Q14 (1 mark, solved), Q15 (1 mark, solved), Q28 (3 marks, solved), Q30 (3 marks, solved)

Set 30/5/1: Q11 (1 mark, solved), Q16 (1 mark, solved), Q22 (2 marks, solved), Q33(a) (5 marks, solved), Q33(b) (5 marks, solved)

Set 30/6/1: Q5 (1 mark, solved), Q14 (1 mark, solved), Q24 (2 marks, solved), Q32(A) (5 marks, solved), Q32(B) (5 marks, solved)

Basic (101 questions, sample papers included)

For 2027

Sample paper 2026-27: Q11 (1 mark, solved), Q17 (1 mark, solved), Q24 (2 marks, solved), Q27(A) (3 marks, solved), Q27(B) (3 marks, solved)

2026

Set 430/1/1: Q9 (1 mark, solved), Q23 (2 marks, solved), Q29 (3 marks, solved)

Set 430/2/1: Q9 (1 mark, solved), Q23 (2 marks, solved), Q29 (3 marks, solved)

Set 430/3/1: Q11 (1 mark, solved), Q23 (2 marks, solved), Q31 (3 marks, solved)

Set 430/4/1: Q4 (1 mark, solved), Q12 (1 mark, solved), Q18 (1 mark, solved), Q24(a) (2 marks, solved), Q24(b) (2 marks, solved), Q26 (3 marks, solved)

Set 430/5/1: Q7 (1 mark, solved), Q15 (1 mark, solved)

Sample paper 2025-26: Q11 (1 mark, solved), Q15 (1 mark, solved), Q16 (1 mark, solved), Q22 (2 marks, solved), Q28 (3 marks, solved)

2025

Set 430/1/1: Q9 (1 mark, solved), Q23 (2 marks, solved), Q29 (3 marks, solved)

Set 430/2/1: Q9 (1 mark, solved), Q23 (2 marks, solved), Q29 (3 marks, solved)

Set 430/3/1: Q11 (1 mark, solved), Q23 (2 marks, solved), Q31 (3 marks, solved)

Set 430/4/1: Q8 (1 mark, solved), Q17 (1 mark, solved), Q24(a) (2 marks, solved), Q24(b) (2 marks, solved), Q31 (3 marks, solved)

Set 430/5/1: Q5 (1 mark, solved), Q14 (1 mark, solved), Q16 (1 mark, solved)

Set 430/6/1: Q8 (1 mark, solved), Q13 (1 mark, solved), Q33(a) (5 marks, solved)

Sample paper 2024-25: Q5 (1 mark, solved), Q12 (1 mark, solved), Q22 (2 marks, solved), Q30(A) (3 marks, solved), Q30(B) (3 marks, solved)

2024

Set 430/1/1: Q19 (1 mark, solved), Q29(A) (3 marks, solved), Q29(B) (3 marks, solved), Q38 (4 marks, solved)

Set 430/2/1: Q16 (1 mark, solved), Q17 (1 mark, solved), Q25 (2 marks, solved), Q27(a) (3 marks, solved), Q27(b) (3 marks, solved)

Set 430/3/1: Q1 (1 mark, solved), Q9 (1 mark, solved), Q33(a) (5 marks, solved), Q33(b) (5 marks, solved)

Set 430/4/1: Q12 (1 mark, solved), Q14 (1 mark, solved), Q17 (1 mark, solved), Q22 (2 marks, solved), Q27(a) (3 marks, solved), Q27(b) (3 marks, solved)

Set 430/5/1: Q13 (1 mark, solved), Q16 (1 mark, solved), Q24 (2 marks, solved), Q29(a) (3 marks, solved), Q29(b) (3 marks, solved)

Sample paper 2023-24: Q9 (1 mark, solved), Q23 (2 marks, solved), Q29 (3 marks, solved)

2023

Set 430/1/1: Q14 (1 mark, solved), Q17 (1 mark, solved), Q18 (1 mark, solved), Q19 (1 mark, solved), Q28 (3 marks, solved)

Set 430/2/1: Q5 (1 mark, solved), Q28(a) (3 marks, solved), Q28(b) (3 marks, solved), Q37 (4 marks, solved)

Set 430/4/1: Q17 (1 mark, solved), Q18 (1 mark, solved), Q19 (1 mark, solved), Q24 (2 marks, solved), Q31(A) (3 marks, solved), Q31(B) (3 marks, solved)

Set 430/5/1: Q2 (1 mark, solved), Q14 (1 mark, solved), Q23 (2 marks, solved), Q31 (3 marks, solved)

Set 430/6/1: Q14 (1 mark, solved), Q25 (2 marks, solved), Q38 (4 marks, solved)

Practice questions on Circles

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
A quadrilateral $A B C D$ is drawn to circumscribe a circle. If $A B = 6$ cm, $B C = 7$ cm and $C D = 4$ cm, then $A D$ is
(a) $3$ cm
(b) $5$ cm
(c) $9$ cm
(d) $17$ cm
Hint
Sums of opposite sides are equal.
Answer
(a) $3$ cm. $A B + C D = A D + B C$ gives $6 + 4 = A D + 7$ , so $A D = 3$ cm.
Q2 · 2 marks · Very short answerBoard level
$P A$ and $P B$ are tangents from $P$ to a circle with centre $O$ and radius $6$ cm. If $∠ A P B = 9 0^{\circ}$ , find $P A$ and $O P$ .
Hint
Find all four angles of $O A P B$ first.
Answer
$∠ O A P = ∠ O B P = 9 0^{\circ}$ and $∠ A O B = 18 0^{\circ} - 9 0^{\circ} = 9 0^{\circ}$ . With $O A = O B$ and $P A = P B$ , $O A P B$ is a square. So $P A = 6$ cm and $O P = 62$ cm.
Q3 · 1 mark · Assertion-reasonBoard level
Two tangents from an external point $P$ touch a circle with centre $O$ at $A$ and $B$ .
Assertion (A): If the tangents are inclined to each other at $8 0^{\circ}$ , then $∠ A O B = 10 0^{\circ}$ .
Reason (R): The angle between the two tangents is equal to the angle subtended by $A B$ at the centre.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Add the angles of quadrilateral $O A P B$ .
Answer
(c). The two right angles at $A$ and $B$ give $∠ A P B + ∠ A O B = 18 0^{\circ}$ , so $∠ A O B = 10 0^{\circ}$ and A is true. The angles are supplementary, not equal, so R is false.
Q4 · 3 marks · Short answerHarder board level
A circle is inscribed in $△ A B C$ in which $A B = A C$ . It touches $B C$ , $C A$ and $A B$ at $D$ , $E$ and $F$ . Prove that $D$ is the midpoint of $B C$ .
Hint
Subtract the equal tangents from $A$ from the equal sides.
Answer
Equal tangents: $A F = A E$ , $B F = B D$ , $C E = C D$ . Since $A B = A C$ , $A B - A F = A C - A E$ , so $B F = C E$ . Hence $B D = B F = C E = C D$ , so $D$ is the midpoint of $B C$ .
Q5 · 3 marks · Short answerHarder board level
The length of a tangent from a point $P$ to a circle is $24$ cm, and the distance from $P$ to the nearest point of the circle is $16$ cm. Find the radius of the circle and the distance of $P$ from the centre.
Hint
Write $O P = r + 16$ and use the right angle at the point of contact.
Answer
$O P = r + 16$ and $O P^{2} = r^{2} + 2 4^{2}$ . So $r^{2} + 32 r + 256 = r^{2} + 576$ , giving $32 r = 320$ , $r = 10$ cm and $O P = 26$ cm. Check: $1 0^{2} + 2 4^{2} = 676 = 2 6^{2}$ .
Q6 · 3 marks · Short answerHarder board level
$T P$ and $T Q$ are tangents from $T$ to a circle with centre $O$ and radius $5$ cm, touching it at $P$ and $Q$ . If $O T = 13$ cm, find $T P$ and the length of the chord $P Q$ .
Hint
$O T$ cuts $P Q$ at right angles at its midpoint $M$ ; use the area of $△ O P T$ two ways.
Answer
$T P = 169 - 25 = 12$ cm. $O T ⊥ P Q$ at $M$ with $P M = M Q$ . Area of $△ O P T$ : $\frac{1}{2} \times 5 \times 12 = \frac{1}{2} \times 13 \times P M$ , so $P M = \frac{60}{13}$ cm and $P Q = \frac{120}{13}$ cm.
Q7 · 3 marks · Short answerHarder board level
Quadrilateral $A B C D$ circumscribes a circle, with $A B = (3 x - 1)$ cm, $B C = (2 x + 3)$ cm, $C D = (x + 6)$ cm and $D A = 12$ cm. Prove the relation you use, then find $x$ and the perimeter of $A B C D$ .
Hint
Write each side as a sum of two tangent lengths to prove $A B + C D = A D + B C$ .
Answer
Equal tangents from each vertex give $A B + C D = A D + B C$ . So $4 x + 5 = 2 x + 15$ , giving $x = 5$ . Sides: $14$ , $13$ , $11$ , $12$ cm; check $14 + 11 = 25 = 12 + 13$ . Perimeter $= 50$ cm.
Q8 · 5 marks · Long answerHarder board level
A circle touches the side $B C$ of $△ A B C$ at $P$ , and touches $A B$ produced at $Q$ and $A C$ produced at $R$ . Prove that $A Q = \frac{1}{2}$ (perimeter of $△ A B C$ ). If $A B = 7$ cm, $B C = 8$ cm and $C A = 9$ cm, find $A Q$ , $B P$ and $C P$ .
Hint
Use equal tangents from $A$ , $B$ and $C$ , and split $B C$ into $B P + P C$ .
Answer
$A Q = A R$ , $B Q = B P$ , $C R = C P$ . Perimeter $= A B + B P + P C + C A = A B + B Q + C A + C R = A Q + A R = 2 A Q$ . Here perimeter $= 24$ cm, so $A Q = 12$ cm, $B P = B Q = 12 - 7 = 5$ cm and $C P = C R = 12 - 9 = 3$ cm (check: $5 + 3 = 8$ ).
Q9 · 3 marks · Short answerHOTS / competency
A circular clock face of radius $9$ cm hangs from a nail $N$ on a wall by a string. The two straight parts of the string, $N A$ and $N B$ , are tangents to the clock's rim, touching it at $A$ and $B$ , and each is $9$ cm long. Find the angle between the two parts of the string and the distance of the nail from the centre $O$ of the clock.
Hint
Look at the shape $O A N B$ : its sides and the angles at $A$ and $B$ .
Answer
$O A = A N = N B = B O = 9$ cm, so $O A N B$ is a rhombus, and $∠ O A N = 9 0^{\circ}$ (tangent ⟂ radius), so it is a square. Hence $∠ A N B = 9 0^{\circ}$ and $O N = 92$ cm.
Q10 · 2 marks · Short answerHOTS / competency
A point $P$ is $10$ cm from the centre of a circle of radius $12$ cm. A student writes: "Tangent length $P T = 1 2^{2} - 1 0^{2} = 44$ cm." Explain what is wrong with this answer.
Hint
Compare $O P$ with the radius before using any formula.
Answer
$O P = 10$ cm is less than the radius $12$ cm, so $P$ lies inside the circle and no tangent can be drawn from it. The formula $P T^{2} = O P^{2} - r^{2}$ would give $100 - 144 < 0$ ; the student swapped the terms to hide this.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What is a tangent to a circle?Show answer
A line that meets the circle at exactly one point, called the point of contact.
2.What is a secant?Show answer
A line that cuts a circle at two distinct points.
3.State Theorem 1 of this chapter.Show answer
The tangent at any point of a circle is perpendicular to the radius through the point of contact.
4.State Theorem 2 of this chapter.Show answer
The lengths of the two tangents drawn from an external point to a circle are equal.
5.How many tangents can be drawn from a point inside, on and outside a circle?Show answer
Inside: none. On: exactly one. Outside: exactly two.
6.If the angle between two tangents from P is $4 0^{\circ}$ , what angle does the chord of contact subtend at the centre?Show answer
$18 0^{\circ} - 4 0^{\circ} = 14 0^{\circ}$ .
7.Which congruence rule proves that tangents from an external point are equal?Show answer
RHS: right angles at the points of contact, equal radii, common hypotenuse OP.
8.A parallelogram circumscribes a circle. What kind of parallelogram must it be?Show answer
A rhombus, because opposite sides sum equally, which forces adjacent sides to be equal.

Best order to study it

Learn the words: secant, tangent, point of contact, and the three positions of a point (inside, on, outside) with the number of tangents for each.
Learn the proof of Theorem 1 (tangent perpendicular to radius) until you can write it with every reason, then practise Pythagoras problems on the tangent triangle, including concentric circles.
Learn the proof of Theorem 2 (equal tangents) by RHS, and note the extra results: OP bisects the angles and is the perpendicular bisector of the chord of contact.
Practise angle chasing with two tangents using the kite OAPB and the isosceles triangles PAB and OAB.
Move to triangles and quadrilaterals around a circle: equal-tangent labelling, AB + CD = AD + BC, right-triangle inradius and the third-tangent perimeter.
Finish with mixed proofs, assertion-reason items and case-based questions under timed conditions.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you write the full proof that a tangent is perpendicular to the radius, with the reason for OQ being longer than OP?
Can you prove that two tangents from an external point are equal, naming all three RHS pairs?
Can you find the tangent length when the radius is 7 cm and the point is 25 cm from the centre? (24 cm)
Can you say why the angle between two tangents and the angle at the centre add to 180 degrees?
Can you find the angle OAB when the angle between the tangents is 70 degrees? (35 degrees)
Can you find the chord of a 13 cm circle that touches a concentric 5 cm circle? (24 cm)
Can you prove that tangents at the ends of a diameter are parallel?
Can you prove that a parallelogram circumscribing a circle is a rhombus?
Can you find the three tangent lengths of a triangle with sides 12, 8 and 10 cm with an incircle? (7, 5 and 3 cm)
Can you find the inradius of a right triangle with legs 6 cm and 8 cm? (2 cm)

Circles: frequently asked questions

Which proofs must I learn for Circles in Class 10?

You must learn two proofs: the tangent at any point is perpendicular to the radius through the point of contact, and the tangents from an external point are equal in length. Both are named in the syllabus as proofs, so write them with Given, To prove, Construction and full reasons.

Are constructions of tangents in the 2026-27 syllabus?

No. Constructions of tangents were removed in the NCERT rationalisation and are not in the 2026-27 syllabus.

Do I need to draw the figure in a proof?

Yes. A neat labelled figure is expected, and in a proof it carries part of the marks along with Given and To prove. Mark the right angles at the points of contact on the figure.

How do I know which side is the hypotenuse in tangent problems?

The right angle is always at the point of contact, between the radius and the tangent. The side opposite it, from the centre to the external point, is the hypotenuse. So the square of the tangent length equals the square of that distance minus the square of the radius.

Can I directly quote results like the angle between tangents being supplementary to the angle at the centre?

In an MCQ or assertion-reason question you can use it directly. In a 2-, 3- or 5-mark answer it is safer to show the two lines of working from the quadrilateral formed by the centre, the two points of contact and the external point, because the examiner looks for the reasons.

Based on: CBSE Mathematics syllabus 2026-27; NCERT Mathematics, Textbook for Class X (Reprint 2026-27)