MathsICSE Class 10 · Chapter 16

Tangent and Secant Properties ICSE Class 10: Theorems, Proofs and Solved Questions

Updated 14 October 2026 · Fact-checked

A tangent is perpendicular to the radius at the point of contact, and the two tangents from an outside point are equal. When chords or secants meet, the products of their segments are equal; for a tangent PT and secant PAB, PT² = PA × PB. The tangent–chord angle equals the angle in the alternate segment.

What Tangent and Secant Properties is about

A tangent touches a circle at exactly one point; a secant cuts it at two points. This chapter gives the length and angle facts that connect tangents, chords and secants.

The CISCE 2027 syllabus covers: the tangent at a point is perpendicular to the radius through it; if two circles touch, the point of contact lies on the line joining their centres; the two tangents from an external point are equal; if two chords intersect internally or externally, the products of their segments are equal; for a tangent and a secant from an external point, the product of the secant's segments equals the square of the tangent; and the alternate segment theorem. Proofs are to be taught.

The CISCE examiners' comments for 2024 and 2025 show where marks go: not using radius ⟂ tangent, not applying the alternate segment theorem, using the wrong segments in $P T^{2} = P A \times P B$ , and in proofs, matching the wrong angles or stating the result without proving it.

Weightage: CISCE publishes no chapter or unit weightage for ICSE Mathematics. Tangent and Secant Properties is in the 3. Geometry (c) Circles (iii) Tangent and Secant Properties part of the 2027 syllabus and can be asked in both Section A and Section B of the 80-mark paper.Source: CISCE ICSE Mathematics syllabus, examination year 2027

Key concepts

Tangent perpendicular to the radius

The tangent at any point of a circle is perpendicular to the radius through the point of contact: if $P T$ touches the circle at $T$ , then $∠ O T P = 9 0^{\circ}$ .

So triangle $O T P$ is right-angled at $T$ and $P T^{2} = O P^{2} - r^{2}$ . Conversely, a line through the end of a radius and perpendicular to it is a tangent.

Worked example

A point $P$ is 13 cm from the centre of a circle of radius 5 cm. Find the length of the tangent from $P$ .

Try it first, then show the solutionSolution

$∠ O T P = 9 0^{\circ}$ (tangent ⟂ radius).
$P T = 1 3^{2} - 5^{2} = 144 = 12$ cm.

Answer: 12 cm

Tangents from an external point are equal

From a point $P$ outside a circle, exactly two tangents $P A$ and $P B$ can be drawn, and $P A = P B$ .

Proof: in $△ O A P$ and $△ O B P$ , $O A = O B$ (radii), $O P$ is common, and $∠ O A P = ∠ O B P = 9 0^{\circ}$ , so the triangles are congruent (RHS) and $P A = P B$ .

Consequences: $O P$ bisects $∠ A P B$ and $∠ A O B$ ; $△ P A B$ is isosceles; and in quadrilateral $O A P B$ , $∠ A P B + ∠ A O B = 18 0^{\circ}$ .

Worked example

$P A$ and $P B$ are tangents to a circle with centre $O$ , and $∠ A P B = 7 0^{\circ}$ . Find $∠ A O B$ and $∠ P A B$ .

Try it first, then show the solutionSolution

$∠ O A P = ∠ O B P = 9 0^{\circ}$ , so $∠ A O B = 36 0^{\circ} - 18 0^{\circ} - 7 0^{\circ} = 11 0^{\circ}$ .
$P A = P B$ , so $∠ P A B = \frac{18 0 ^{\circ} - 7 0 ^{\circ}}{2} = 5 5^{\circ}$ .

Answer: ∠AOB = 110°, ∠PAB = 55°

Touching circles

If two circles touch, the point of contact lies on the straight line joining their centres.

Touching externally: distance between centres $= R + r$ .
Touching internally: distance between centres $= R - r$ .

The common tangent at the point of contact is perpendicular to the line of centres.

Worked example

Two circles of radii 5 cm and 3 cm touch each other. Find the distance between their centres if they touch (a) externally, (b) internally.

Try it first, then show the solutionSolution

(a) $5 + 3 = 8$ cm.
(b) $5 - 3 = 2$ cm.

Answer: (a) 8 cm (b) 2 cm

Chords intersecting inside the circle

If chords $A B$ and $C D$ meet at $P$ inside the circle, then
$P A \times P B = P C \times P D .$

Proof: in $△ P A C$ and $△ P D B$ , $∠ A P C = ∠ D P B$ (vertically opposite) and $∠ C A B = ∠ C D B$ (same segment), so they are similar (AA) and $\frac{P A}{P D} = \frac{P C}{P B}$ .

Worked example

Chords $A B$ and $C D$ meet at $P$ inside a circle. $A P = 4$ cm, $P B = 6$ cm and $C P = 3$ cm. Find $P D$ .

Try it first, then show the solutionSolution

$4 \times 6 = 3 \times P D$ .
$P D = 8$ cm.

Answer: PD = 8 cm

Chords (secants) intersecting outside the circle

If chords $A B$ and $C D$ , produced, meet at $P$ outside the circle (so $P A B$ and $P C D$ are secants), then
$P A \times P B = P C \times P D,$
where each length is measured **from $P$ **: $P B = P A + A B$ .

The proof uses $△ P A C \sim △ P D B$ : $∠ P$ is common and $∠ P A C = ∠ P D B$ (exterior angle of cyclic quadrilateral $A B D C$ ).

Worked example

$P A B$ and $P C D$ are secants from an external point $P$ . $P A = 4$ cm, $A B = 5$ cm and $P C = 3$ cm. Find $P D$ and $C D$ .

Try it first, then show the solutionSolution

$P B = 4 + 5 = 9$ cm.
$4 \times 9 = 3 \times P D$ , so $P D = 12$ cm.
$C D = 12 - 3 = 9$ cm.

Answer: PD = 12 cm, CD = 9 cm

Tangent and secant from the same point

If $P T$ is a tangent and $P A B$ a secant from the same external point $P$ , then
$P T^{2} = P A \times P B .$

Proof: in $△ P T A$ and $△ P B T$ , $∠ P$ is common and $∠ P T A = ∠ P B T$ (alternate segment theorem), so they are similar (AA). Hence $\frac{P T}{P B} = \frac{P A}{P T}$ , giving $P T^{2} = P A \times P B$ .

If the secant passes through the centre, $P A = O P - r$ and $P B = O P + r$ , so $P T^{2} = O P^{2} - r^{2}$ , the same as Pythagoras.

Worked example

From $P$ , a tangent $P T$ and a secant $P A B$ are drawn. $P A = 4$ cm and $P B = 9$ cm. Find $P T$ .

Try it first, then show the solutionSolution

$P T^{2} = 4 \times 9 = 36$ .
$P T = 6$ cm.

Answer: 6 cm

The alternate segment theorem

If a line touches a circle and a chord is drawn from the point of contact, the angle between the tangent and the chord equals the angle in the alternate segment (the segment on the other side of the chord).

If $T A$ is tangent at $A$ and $A B$ is a chord, then $∠ T A B = ∠ A C B$ for any $C$ on the arc on the other side of $A B$ from $T$ .

The proof draws the diameter $A D$ : $∠ T A D = 9 0^{\circ}$ and $∠ A B D = 9 0^{\circ}$ , then compares angles in the same segment.

Worked example

$T A$ is a tangent at $A$ and $A B$ a chord with $∠ T A B = 5 8^{\circ}$ . $C$ is on the major arc. Find $∠ A C B$ and the angle subtended by $A B$ at the centre.

Try it first, then show the solutionSolution

$∠ A C B = ∠ T A B = 5 8^{\circ}$ (alternate segment).
$∠ A O B = 2 \times 5 8^{\circ} = 11 6^{\circ}$ (angle at the centre).

Answer: ∠ACB = 58°; ∠AOB = 116°

Maths in real life: Tangent and Secant Properties

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.

Tangent length with Pythagoras

Suppose a circular pond has radius 15 m, and a lamp post $P$ stands 39 m from its centre. Straight paths run from $P$ to touch the pond's edge.

How long is each path?

Try it first, then show the solutionSolution

The path is a tangent, so it is perpendicular to the radius at the contact point.
Length $= 3 9^{2} - 1 5^{2} = 1296 = 36$ m.

Answer: 36 m

Memory hook: Tangent, radius and distance to the centre make a right triangle.

Touching circles

Imagine two gear wheels of radii 9 cm and 5 cm meshing so that their rims just touch.

How far apart are the axles?

Try it first, then show the solutionSolution

The point of contact lies on the line of centres.
They touch externally: distance $= 9 + 5 = 14$ cm.

Answer: 14 cm

Memory hook: Touching outside: add the radii.

Intersecting chords

Suppose two straight cables cross inside a circular window frame. One cable is split into pieces of 30 cm and 40 cm at the crossing, and one piece of the other is 24 cm.

How long is the other piece of the second cable?

Try it first, then show the solutionSolution

$30 \times 40 = 24 \times x$ .
$x = 50$ cm.

Answer: 50 cm

Memory hook: Crossing chords: products of the pieces match.

Topics in this chapter

Tangent and Secant Properties formula sheet

Tangent and radius: $O T ⊥ P T, P T^{2} = O P^{2} - r^{2}$; Right angle at the point of contact.
Equal tangents: $P A = P B, ∠ A P B + ∠ A O B = 18 0^{\circ}$; OP bisects both angles.
Touching circles: $d = R + r (external), d = R - r (internal)$; Contact point is on the line of centres.
Intersecting chords: $P A \times P B = P C \times P D$; Inside or outside the circle; measure from P.
Tangent–secant: $P T^{2} = P A \times P B$; PB is the whole secant from P.
Alternate segment: $∠ T A B = ∠ A C B$; C in the segment on the other side of AB.
Circle inside a quadrilateral: $A B + C D = A D + B C$; From equal tangents at the four corners.

How the board asks Tangent and Secant Properties

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

Question type	How it is asked	Method
MCQ: tangent length or an angle	Section A, Question 1: one property, often equal tangents with an isosceles triangle.	Mark the right angles and equal tangents first.
Lengths from chords, secants and tangents	3 or 4 marks: find a segment or a radius.	Pick the right product rule; measure every length from the meeting point.
Angles with tangents	A figure with a tangent, chords and the centre; find several angles with reasons.	Radius ⟂ tangent, alternate segment, then circle angle properties.
Prove a property	Prove equal tangents, PT² = PA × PB, or a pair of similar triangles first.	Name the equal angles with three letters and correct reasons, then write the proportion.
Incircle and touching circles	Find tangent lengths of an incircle or the distance between centres.	Equal tangents from each vertex; R ± r for touching circles.

Tricks to remember Tangent and Secant Properties

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

Radius meets tangent at 90
Whenever a tangent appears, draw the radius to the contact point and mark the square corner.
Use it when: Finding the first angle or length.
Outside × whole
For a secant from P: (outside part) × (whole secant). For a tangent: tangent × tangent.
Use it when: PA × PB = PC × PD and PT² = PA × PB.
Ice-cream cone
Two tangents from P look like a cone: equal sides, so the triangle with the chord is isosceles.
Use it when: Equal-tangent angle questions.
Tangent angle jumps across
The angle between tangent and chord jumps across the chord to the far arc.
Use it when: Alternate segment theorem.

Tangent and Secant Properties: board-level solved questions

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

Example 1: Tangent length · 1 mark · MCQBoard level

The length of the tangent from a point 25 cm from the centre of a circle of radius 7 cm is:
(a) 18 cm
(b) 24 cm
(c) 26 cm
(d) 32 cm

Try it first, then show the solutionSolution

$625 - 49 = 576 = 24$ .

Answer: (b)

Example 2: Equal tangents: assertion and reason · 1 mark · Assertion-reasonBoard level

Assertion (A): If $P A$ and $P B$ are tangents from an external point $P$ , then $∠ P A B = ∠ P B A$ .
Reason (R): The lengths of the two tangents from an external point are equal.
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).

Try it first, then show the solutionSolution

$P A = P B$ , so $△ P A B$ is isosceles and its base angles are equal.

Answer: (c)

Example 3: Chords with a quadratic · 3 marks · Short answerBoard level

Chords $A B$ and $C D$ of a circle meet at $P$ inside it. $A P = 6$ cm, $P B = 4$ cm and $C D = 11$ cm. Find $C P$ and $P D$ .

Try it first, then show the solutionSolution

Let $C P = x$ , $P D = 11 - x$ . $6 \times 4 = x (11 - x)$ .
$x^{2} - 11 x + 24 = 0$ , $(x - 3) (x - 8) = 0$ .
$C P = 3$ cm and $P D = 8$ cm (or the other way round).

Answer: 3 cm and 8 cm

Example 4: Radius from a tangent · 4 marks · Short answerHarder board level

$A B$ is a diameter of a circle. $A B$ is produced beyond $B$ to $P$ , and $P T$ is a tangent to the circle at $T$ . $P T = 12$ cm and $P B = 8$ cm. Find (a) $P A$ , (b) the radius, (c) check your answer with Pythagoras.

Try it first, then show the solutionSolution

(a) $P T^{2} = P B \times P A$ : $144 = 8 \times P A$ , so $P A = 18$ cm.
(b) $A B = P A - P B = 10$ cm, so the radius is 5 cm.
(c) $O P = P B + r = 13$ cm, $O T = 5$ cm, $∠ O T P = 9 0^{\circ}$ : $1 3^{2} - 5^{2} = 12$ ✓

Answer: (a) 18 cm (b) 5 cm

Example 5: Two tangents and the alternate segment · 4 marks · Short answerHarder board level

$T A$ and $T B$ are tangents from $T$ to a circle with centre $O$ ; $C$ is a point on the major arc $A B$ . $∠ A T B = 5 0^{\circ}$ . Find (a) $∠ T A B$ , (b) $∠ A C B$ , (c) $∠ A O B$ , (d) $∠ O A B$ .

Try it first, then show the solutionSolution

(a) $T A = T B$ , so $∠ T A B = \frac{18 0 ^{\circ} - 5 0 ^{\circ}}{2} = 6 5^{\circ}$ .
(b) $∠ A C B = ∠ T A B = 6 5^{\circ}$ (alternate segment).
(c) $∠ A O B = 2 \times 6 5^{\circ} = 13 0^{\circ}$ (angle at the centre); check: $13 0^{\circ} + 5 0^{\circ} = 18 0^{\circ}$ .
(d) $∠ O A B = 9 0^{\circ} - 6 5^{\circ} = 2 5^{\circ}$ (radius ⟂ tangent).

Answer: (a) 65° (b) 65° (c) 130° (d) 25°

Example 6: Touching circles · 3 marks · Long answerBoard level

Two circles with centres $A$ and $B$ and radii 7 cm and 4 cm touch each other at $P$ . (a) Find $A B$ if they touch externally, and the ratio $A P : P B$ . (b) Find $A B$ if they touch internally.

Try it first, then show the solutionSolution

(a) $P$ lies on $A B$ (line of centres): $A B = 7 + 4 = 11$ cm; $A P : P B = 7 : 4$ .
(b) $A B = 7 - 4 = 3$ cm.

Answer: (a) 11 cm, 7 : 4 (b) 3 cm

Example 7: Prove PT² = PA × PB · 4 marks · Long answerHarder board level

$P T$ is a tangent at $T$ and $P A B$ is a secant of a circle. (a) Prove that $△ P T A \sim △ P B T$ . (b) Hence prove that $P T^{2} = P A \times P B$ . (c) Find $P T$ if $P A = 5$ cm and $A B = 15$ cm.

Try it first, then show the solutionSolution

(a) $∠ T P A = ∠ B P T$ (common) and $∠ P T A = ∠ P B T$ (alternate segment, chord $T A$ ). So $△ P T A \sim △ P B T$ (AA).
(b) $\frac{P T}{P B} = \frac{P A}{P T}$ , so $P T^{2} = P A \times P B$ .
(c) $P B = 20$ cm; $P T^{2} = 5 \times 20 = 100$ , $P T = 10$ cm.

Answer: (c) PT = 10 cm

Example 8: Incircle of a triangle and a circumscribed quadrilateral · 5 marks · Long answerHOTS / competency

(a) The incircle of triangle $A B C$ touches $B C$ , $C A$ and $A B$ at $D$ , $E$ and $F$ . $A B = 12$ cm, $B C = 8$ cm and $C A = 10$ cm. Find $A F$ , $B D$ and $C E$ . (b) A circle touches all four sides of quadrilateral $P QR S$ . Prove that $P Q + R S = P S + QR$ , and find $P S$ if $P Q = 9$ cm, $QR = 8$ cm and $R S = 6$ cm.

Try it first, then show the solutionSolution

(a) Let $A F = A E = x$ , $B F = B D = y$ , $C D = C E = z$ (equal tangents). $x + y = 12$ , $y + z = 8$ , $z + x = 10$ .
Adding: $x + y + z = 15$ , so $x = 7$ , $y = 5$ , $z = 3$ : $A F = 7$ , $B D = 5$ , $C E = 3$ cm.
(b) Tangents from each vertex are equal: $P$ gives $a$ , $Q$ gives $b$ , $R$ gives $c$ , $S$ gives $d$ .
$P Q + R S = (a + b) + (c + d)$ and $P S + QR = (a + d) + (b + c)$ : equal.
$P S = 9 + 6 - 8 = 7$ cm.

Answer: (a) 7 cm, 5 cm, 3 cm (b) PS = 7 cm

Case-study practice (Section B style)

Read the situation, then answer every part. In the board paper this is a 10-mark question split into 3 + 3 + 4 marks.

Paths to a circular pond

A circular pond has centre $O$ and radius 15 m. A lamp post $P$ is 39 m from $O$ . Two straight paths $P A$ and $P B$ touch the pond at $A$ and $B$ , and a third straight path from $P$ passes through $O$ , meeting the edge at $C$ (near $P$ ) and $D$ (far side).

(i) · 3 marks
Find the length of path $P A$ .
Answer
Answer: 36 m
$∠ O A P = 9 0^{\circ}$ (tangent ⟂ radius). $P A = 3 9^{2} - 1 5^{2} = 36$ m.
(ii) · 3 marks
Explain why $P B = P A$ and why $O P$ bisects $∠ A P B$ .
Answer
Answer: Congruent triangles (RHS)
In $△ O A P$ and $△ O B P$ : $O A = O B$ (radii), $O P$ common, right angles at $A$ and $B$ . So they are congruent (RHS), giving $P A = P B$ and $∠ O P A = ∠ O P B$ .
(iii) · 4 marks
Find $P C$ and $P D$ , and verify that $P A^{2} = P C \times P D$ .
Answer
Answer: PC = 24 m, PD = 54 m; 1296 = 1296
$P C = 39 - 15 = 24$ m, $P D = 39 + 15 = 54$ m. $P C \times P D = 1296 = 3 6^{2} = P A^{2}$ ✓ (tangent–secant property).

Two pulleys and a belt

Two pulleys with centres $A$ and $B$ and radii 9 cm and 5 cm touch each other externally. A straight belt section $P Q$ touches both, at $P$ on the larger and $Q$ on the smaller, on the same side of $A B$ .

(i) · 3 marks
Find $A B$ and state where the point of contact lies.
Answer
Answer: 14 cm; on AB
Touching externally: $A B = 9 + 5 = 14$ cm, and the contact point lies on the line of centres $A B$ .
(ii) · 3 marks
Explain why $A P$ and $B Q$ are both perpendicular to $P Q$ , and hence parallel.
Answer
Answer: Radius ⟂ tangent
$P Q$ is a tangent to each circle, so it is perpendicular to the radius at each point of contact. Two lines perpendicular to the same line are parallel.
(iii) · 4 marks
Find the length $P Q$ correct to 2 decimal places.
Answer
Answer: 13.42 cm
Draw $B M ∥ P Q$ meeting $A P$ at $M$ . Then $P M B Q$ is a rectangle, $A M = 9 - 5 = 4$ cm, $∠ A M B = 9 0^{\circ}$ . $P Q = M B = 1 4^{2} - 4^{2} = 180 \approx 13.42$ cm.

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.

A point $P$ is 17 cm from the centre of a circle of radius 8 cm. Find the length of the tangent from $P$ . (3 marks)

Step	Marks
$∠ O T P = 9 0^{\circ}$ (tangent ⟂ radius)	1
$P T^{2} = 1 7^{2} - 8^{2} = 225$	1
$P T = 15$ cm	1

$X B Y$ is the tangent at $B$ to a circle through $A$ , $B$ , $C$ , with $X$ on the same side as $A$ . $∠ A B X = 6 5^{\circ}$ and $∠ C B Y = 4 5^{\circ}$ . Find $∠ A C B$ , $∠ B A C$ and $∠ A B C$ . (4 marks)

Step	Marks
$∠ A C B = ∠ A B X = 6 5^{\circ}$ (alternate segment)	1
$∠ B A C = ∠ C B Y = 4 5^{\circ}$ (alternate segment)	1
$∠ A B C = 18 0^{\circ} - 6 5^{\circ} - 4 5^{\circ}$ (angles on a straight line)	1
$= 7 0^{\circ}$ (check: $65 + 45 + 70 = 180$ )	1

$P A B$ and $P C D$ are secants of a circle from an external point $P$ , and $P T$ is a tangent. $P A = 6$ cm, $A B = 10$ cm and $P C = 8$ cm. Find $C D$ and $P T$ . (5 marks)

Step	Marks
$P B = 16$ cm	1
$P A \times P B = P C \times P D$ : $96 = 8 \times P D$	1
$P D = 12$ cm, so $C D = 4$ cm	1
$P T^{2} = P A \times P B = 96$	1
$P T = 96 \approx 9.80$ cm	1

Where students lose marks in Tangent and Secant Properties

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

Mistake: Forgetting radius ⟂ tangent
CISCE examiners' comments, 2024: candidates did not use the fact that the radius is perpendicular to the tangent at Q, so ∠OQM came out wrong.
Do this: Draw the radius to every point of contact and mark 90° before anything else.
Usually costs: 1 mark, plus later parts that depend on it.
Mistake: Missing the alternate segment theorem
CISCE examiners' comments, 2024: candidates could not apply the alternate segment theorem to get ∠QNM = 42° and ∠SNM = 25°, and errors carried into later parts.
Do this: When a chord starts at a point of contact, look across the chord for the equal angle.
Usually costs: 2 to 3 marks in a 4-mark part.
Mistake: Wrong segments in PT² = PA × PB
CISCE examiners' comments, 2024: candidates used PQ² = PO × PB or PQ² = AB × BP, and some subtracted the radius 9 instead of the diameter 18 from PA.
Wrong working
PQ² = PO × PB
Right working
PQ² = PB × PA, both measured from P (PB = PA − diameter)
Do this: Both factors start at P: the near part and the whole secant.
Usually costs: 2 to 3 marks.
Mistake: Ignoring equal tangents
CISCE examiners' comments, 2024: in an MCQ, candidates did not use equal tangents from P, so missed that triangle PST is isosceles; some took a chord as a diameter.
Do this: Two tangents from one point: mark them equal and the base angles equal.
Usually costs: 1 mark.
Mistake: Proofs with vague angles or a skipped step
CISCE examiners' comments, 2025: in proving △PTR ∼ △PST, candidates wrote ∠T = ∠S instead of ∠PTR = ∠PST, wrote proportional sides incorrectly, and stated PT² = PR × PS without proving it from the proportion.
Do this: Name each angle with three letters, give the reason, write the proportion, then cross-multiply.
Usually costs: 1 to 2 marks in a 3-mark part.

Practice questions on Tangent and Secant Properties

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
The angle between a tangent to a circle and the radius at the point of contact is:
(a) $4 5^{\circ}$
(b) $6 0^{\circ}$
(c) $9 0^{\circ}$
(d) $18 0^{\circ}$
Hint
Basic property.
Answer
(c)
Q2 · 1 mark · MCQBoard level
Two circles of radii 6 cm and 4 cm touch internally. The distance between their centres is:
(a) 2 cm
(b) 4 cm
(c) 10 cm
(d) 24 cm
Hint
R − r.
Answer
(a)
Q3 · 1 mark · MCQHarder board level
$P A$ and $P B$ are tangents to a circle of radius 5 cm with centre $O$ , and $∠ A P B = 6 0^{\circ}$ . Then $O P$ is:
(a) 5 cm
(b) 10 cm
(c) $53$ cm
(d) $103$ cm
Hint
∠OPA = 30° and sin 30° = OA/OP.
Answer
(b)
Q4 · 1 mark · Assertion-reasonBoard level
Assertion (A): If chords $A B$ and $C D$ meet at $P$ inside a circle, then $P A \times P B = P C \times P D$ .
Reason (R): Triangles $P A C$ and $P D B$ are similar.
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
Hint
How is the chord property proved?
Answer
(c)
Q5 · 4 marks · Short answerHarder board level
$P T$ is a tangent at $T$ , and a secant from $P$ passes through the centre $O$ , meeting the circle at $A$ (nearer) and $B$ . $P T = 8$ cm and $P A = 4$ cm. Find $P B$ and the radius.
Hint
AB is a diameter.
Answer
$64 = 4 \times P B$ , $P B = 16$ cm. $A B = 12$ cm, radius 6 cm.
Q6 · 4 marks · Short answerHarder board level
$T A$ and $T B$ are tangents to a circle with centre $O$ . $∠ A O B = 3 x^{\circ}$ and $∠ A T B = (x + 20)^{\circ}$ . (a) Find $x$ . (b) Show that $△ T A B$ is equilateral. (c) If the radius is 4 cm, find $T A$ to 2 decimal places.
Hint
∠AOB + ∠ATB = 180°.
Answer
(a) $4 x + 20 = 180$ , $x = 40$ . (b) $∠ A T B = 6 0^{\circ}$ and $T A = T B$ , so base angles are $6 0^{\circ}$ : equilateral. (c) $∠ O T A = 3 0^{\circ}$ , $T A = \frac{4}{t a n 3 0 ^{\circ}} = 43 \approx 6.93$ cm.
Q7 · 3 marks · Long answerBoard level
Chords $A B$ and $C D$ meet at $P$ inside a circle. $A P = 5$ cm, $P B = 8$ cm and $C P = 4$ cm. Find $P D$ and $C D$ .
Hint
Products of the pieces.
Answer
$P D = \frac{40}{4} = 10$ cm; $C D = 14$ cm.
Q8 · 4 marks · Long answerHarder board level
$T A$ is a tangent at $A$ and $A B$ a chord with $∠ T A B = 7 0^{\circ}$ . $C$ is on the major arc and $D$ on the minor arc $A B$ . Find $∠ A C B$ , $∠ A D B$ and $∠ A O B$ .
Hint
Alternate segment, then cyclic quadrilateral.
Answer
$∠ A C B = 7 0^{\circ}$ (alternate segment); $∠ A D B = 11 0^{\circ}$ ( $A C B D$ cyclic); $∠ A O B = 14 0^{\circ}$ .
Q9 · 4 marks · Long answerHOTS / competency
Prove that the two tangents from an external point $P$ to a circle with centre $O$ are equal and that $O P$ is the perpendicular bisector of the chord of contact $A B$ .
Hint
RHS congruence, then the two equal-distance loci.
Answer
$△ O A P ≅ △ O B P$ (RHS: right angles at $A$ and $B$ , $O A = O B$ , $O P$ common), so $P A = P B$ . Now $O$ and $P$ are each equidistant from $A$ and $B$ , so both lie on the perpendicular bisector of $A B$ ; hence $O P$ is that bisector.
Q10 · 4 marks · Long answerHarder board level
A circle touches the sides $A B$ , $B C$ , $C D$ , $D A$ of quadrilateral $A B C D$ at $P$ , $Q$ , $R$ , $S$ . $A P = 6$ cm, $A B = 10$ cm, $B C = 7$ cm and $C D = 6$ cm. Find $B Q$ , $C R$ , $D S$ and $A D$ .
Hint
Walk round the corners using equal tangents.
Answer
$B P = B Q = 4$ , $C Q = C R = 3$ , $D R = D S = 3$ , $A S = A P = 6$ . $A D = 6 + 3 = 9$ cm (check: $10 + 6 = 9 + 7$ ).
Q11 · 1 mark · MCQHOTS / competency
From an external point $P$ , a tangent $P T = 6$ cm and a secant $P A B$ with $P A = 3$ cm are drawn. The chord $A B$ is:
(a) 3 cm
(b) 6 cm
(c) 9 cm
(d) 12 cm
Hint
Find PB first.
Answer
(c): $P B = 36 \div 3 = 12$ , $A B = 9$ .
Q12 · 5 marks · Long answerHOTS / competency
The tangent at $T$ to a circle meets the chord $B A$ produced at $P$ (with $A$ between $P$ and $B$ ). (a) Prove that $∠ P T A = ∠ T B A$ . (b) Hence prove $△ P T A \sim △ P B T$ . (c) If $P A = 2$ cm and $A B = 6$ cm, find $P T$ and the ratio of the areas of $△ P T A$ and $△ P B T$ .
Hint
Alternate segment, then AA.
Answer
(a) $∠ P T A$ is the angle between tangent $P T$ and chord $T A$ ; it equals $∠ T B A$ in the alternate segment. (b) $∠ P$ common and $∠ P T A = ∠ P B T$ : AA. (c) $P T^{2} = 2 \times 8 = 16$ , $P T = 4$ cm. Ratio of sides $\frac{P T}{P B} = \frac{4}{8}$ , so areas $1 : 4$ .

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.Angle between a tangent and the radius at the point of contact?Show answer
90°
2.How many tangents from an external point, and how long?Show answer
Two, equal in length.
3.Distance between centres of circles touching externally?Show answer
R + r
4.Chords AB, CD meet at P. Relation?Show answer
PA × PB = PC × PD
5.Tangent PT and secant PAB. Relation?Show answer
PT² = PA × PB
6.State the alternate segment theorem.Show answer
The angle between a tangent and a chord through the point of contact equals the angle in the alternate segment.
7.∠APB = 60° between two tangents. ∠AOB?Show answer
120°

Best order to study it

Tangent ⟂ radius and tangent length by Pythagoras.
Equal tangents (with proof) and the angle facts that follow.
Touching circles.
Intersecting chords inside and outside.
Tangent–secant property with proof.
Alternate segment theorem; mixed angle questions with reasons.

Quick revision checklist

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

Can you prove that tangents from an external point are equal?
Do you measure secant lengths from P?
Can you prove PT² = PA × PB using similar triangles?
Can you spot the alternate segment angle?
Do you use R + r and R − r correctly?
Do you give reasons for every angle?

Tangent and Secant Properties: frequently asked questions

Which tangent proofs are needed for ICSE 2027?

The syllabus note says proofs of these theorems are to be taught unless specified otherwise, so practise proofs of the tangent–radius property, equal tangents, the chord products and PT² = PA × PB.

What is the difference between a tangent and a secant?

A tangent meets the circle at exactly one point; a secant cuts it at two points.

In PA × PB, is PB the whole secant?

Yes. Both lengths are measured from the external point P: PA to the nearer point, PB to the farther point.

What is the alternate segment?

The segment of the circle on the opposite side of the chord from the angle between the tangent and the chord.

Can I use Pythagoras instead of PT² = PA × PB?

When the secant passes through the centre, both give the same answer. Otherwise use the tangent–secant property.

Based on: CISCE ICSE 2027 Mathematics syllabus; CISCE ICSE 2027 Mathematics specimen question paper and marking scheme; CISCE Analysis of Pupil Performance, ICSE Mathematics, 2025; CISCE Analysis of Pupil Performance, ICSE Mathematics, 2024