MathsCBSE Class 10 · Circles

Solving Problems Using Tangent Properties, Class 10

Updated 13 October 2026 · Fact-checked

Solving problems using tangent properties means using two facts, that a tangent is perpendicular to the radius at the point of contact and that tangents from an external point are equal, together with Pythagoras and angle sums. The board asks you to find tangent lengths, radii, distances, angles between tangents and sides of figures drawn around circles.

Radius meets tangent at 90°, so OAP is a right triangle: use PA² = OP² − r², and remember PA = PB.

Illustration drawn for QuestionBuddy.

Understand Solving Problems Using Tangent Properties

Almost every numerical in this chapter is solved with just two theorems and a few tools you already know. The first theorem says the tangent at a point is perpendicular to the radius there. The second says the two tangents from an external point are equal. The skill is spotting which one a question needs and turning the figure into a triangle you can work with.

The right triangle. Join the centre $O$ to the external point $P$ and to the point of contact $T$ . Then $∠ O T P = 9 0^{\circ}$ , so $△ O T P$ is right-angled with hypotenuse $O P$ . This gives $P T^{2} = O P^{2} - r^{2}$ . Any one of the three lengths can be found from the other two. If an angle is given, trigonometric ratios in the same triangle work too, for example $tan ∠ O P T = \frac{r}{P T}$ .

The kite. With two tangents $P A$ and $P B$ , the figure $O A P B$ has two right angles, so $∠ A P B + ∠ A O B = 18 0^{\circ}$ . Triangles $O A P$ and $O B P$ are congruent, so $O P$ bisects $∠ A P B$ and $∠ A O B$ , and $O P$ is the perpendicular bisector of the chord $A B$ . Triangle $P A B$ is isosceles because $P A = P B$ .

Equal tangents around a figure. When a circle touches all sides of a triangle or quadrilateral, every vertex is an external point. Label the two tangents from each vertex with the same letter, write one equation per side, and solve. The detailed proofs for these figures sit on their own page; here you use the results to get numbers.

Conditions. The right angle exists only at the point of contact and only with the radius drawn to that point. Equal tangents need the same external point and the same circle. The point must be outside the circle; from a point on the circle there is one tangent, and from inside there is none. Always write the reason, such as (tangent $⊥$ radius) or (tangents from an external point), next to the line where you use it, because the board gives step marks for reasons.

Solving Problems Using Tangent Properties: formulas to remember

Tangent length: $P T^{2} = O P^{2} - r^{2}$; Right angle at T. OP is the hypotenuse, never a leg.
Equal tangents: $P A = P B$; Same external point, same circle.
Angle between tangents: $∠ A P B + ∠ A O B = 18 0^{\circ}$; From quadrilateral OAPB with two right angles.
Half-angle in the right triangle: $∠ A P O = \frac{1}{2} ∠ A P B, tan ∠ A P O = \frac{r}{P A}$; OP bisects the angle between the tangents.
Chord of contact: $A B ⊥ O P, A M = M B = \frac{P A \cdot r}{O P}$; M is the foot on OP. Comes from equating two expressions for area of triangle OAP.
Chord of larger concentric circle touching the smaller: $A B = 2 R^{2} - r^{2}$; Bisected at the point of contact.
Third tangent perimeter: $P C + C D + D P = 2 P A$; CD touches the circle between tangents PA and PB.
Quadrilateral circumscribed about a circle: $A B + C D = A D + B C$; Sums of opposite sides are equal. Comes from the four pairs of equal tangents from the vertices.
Incircle of a right triangle (right angle at B): $r = \frac{A B + B C - A C}{2}$; The radii to the two legs and the vertex B form a square, so the tangents from B equal r.

How to solve Solving Problems Using Tangent Properties questions

Use this routine on any length or angle question with tangents.

1Describe or sketch the figure and mark the centre, every point of contact and every external point.
2Join the centre to each point of contact and mark $9 0^{\circ}$ there, writing (tangent $⊥$ radius).
3From each external point, mark the two tangents as equal with the same letter, writing (tangents from an external point).
4Choose the tool: Pythagoras in $△ O T P$ for lengths, angle sum in $O A P B$ or isosceles triangles for angles, side equations for figures around the circle.
5Convert disguised data: distance to the circle plus $r$ gives $O P$ ; diameter halved gives $r$ .
6Solve, then check by substituting back (for example $r^{2} + P T^{2} = O P^{2}$ ) and write units.

Solving Problems Using Tangent Properties in real life

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 from an external point

Imagine a circular water tank of radius 9 m stands in your school yard. You stand at a point P on the playground and look along the two straight edges of your view that just graze the tank wall at A and B. You are 6 m from the nearest point of the tank wall.

Find the length of each line of sight PA and the length of the chord AB joining the two grazing points.

Try it first, then show the solutionSolution

$O P = 6 + 9 = 15$ m (distance to wall plus radius).
$∠ O A P = 9 0^{\circ}$ (tangent $⊥$ radius).
$P A = 1 5^{2} - 9^{2} = 144 = 12$ m.
$P B = P A = 12$ m (tangents from an external point).
$O P$ is the perpendicular bisector of $A B$ ; let it meet $A B$ at $M$ .
Area of $△ O A P$ : $\frac{1}{2} \cdot 9 \cdot 12 = \frac{1}{2} \cdot 15 \cdot A M$ , so $A M = 7.2$ m.
$A B = 2 \times 7.2 = 14.4$ m.

Answer: PA = PB = 12 m and AB = 14.4 m

Memory hook: Look past a tank: your sight line, the radius and your distance to the centre make a right triangle.

Angle between two tangents

Suppose your housing society has a circular flower bed with centre O and radius 7 m. From the gate P, two straight paths are laid so that each just touches the edge of the bed, at A and B. The angle between the two paths at the gate is $6 0^{\circ}$ .

Find the distance of the gate from the centre, the length of each path, the area of the kite-shaped plot OAPB, and the length of a straight hosepipe laid along the chord AB.

Try it first, then show the solutionSolution

$∠ O A P = ∠ O B P = 9 0^{\circ}$ (tangent $⊥$ radius).
$△ O A P ≅ △ O B P$ (RHS), so $O P$ bisects $∠ A P B$ and $∠ O P A = 3 0^{\circ}$ .
$sin 3 0^{\circ} = \frac{O A}{O P}$ , so $O P = \frac{7}{1/2} = 14$ m.
$P A = 1 4^{2} - 7^{2} = 147 = 73$ m, and $P B = P A$ (tangents from an external point).
Area of $O A P B = 2 \times \frac{1}{2} \times 7 \times 73 = 493$ m².
$P A = P B$ and $∠ A P B = 6 0^{\circ}$ , so $△ P A B$ is equilateral and $A B = 73$ m.
Check: $O P$ meets $A B$ at $M$ with $A M = \frac{P A \cdot O A}{O P} = \frac{49 3}{14} = \frac{7 3}{2}$ , so $A B = 2 A M = 73$ m.

Answer: OP = 14 m, PA = PB = 7√3 m, area of OAPB = 49√3 m², AB = 7√3 m

Memory hook: Two garden paths hugging a round bed at 60° put the gate at twice the radius from the centre.

Tricks to remember Solving Problems Using Tangent Properties

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

Hypotenuse goes home
The longest side always ends at the centre and the outside point: $O P$ is the hypotenuse.
Use it when: Stops you writing $P T^{2} = O P^{2} + r^{2}$ .
Kite adds to 180
Two right angles in the kite $O A P B$ leave $18 0^{\circ}$ for the other two.
Use it when: Instant MCQ on angle between tangents and angle at centre.
Same vertex, same letter
Give both tangents from a vertex one letter; then every side is a sum of two letters.
Use it when: Triangles and quadrilaterals drawn around a circle.

Solving Problems Using Tangent Properties: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Angle between tangents from the angle at the centre · 2 marks · Very short answerBoard level

$P A$ and $P B$ are tangents from an external point $P$ to a circle with centre $O$ . If $∠ P A B = 5 5^{\circ}$ , find $∠ A O B$ and $∠ O A B$ .

Try it first, then show the solutionSolution

$P A = P B$ (tangents from an external point), so $∠ P B A = ∠ P A B = 5 5^{\circ}$ . (½)
$∠ A P B = 18 0^{\circ} - 11 0^{\circ} = 7 0^{\circ}$ . (½)
$∠ A O B = 18 0^{\circ} - 7 0^{\circ} = 11 0^{\circ}$ (quadrilateral $O A P B$ has two right angles). (½)
$∠ O A B = 9 0^{\circ} - 5 5^{\circ} = 3 5^{\circ}$ (since $∠ O A P = 9 0^{\circ}$ ). (½)

Answer: ∠AOB = 110° and ∠OAB = 35°

Example 2: Radius from a disguised distance · 3 marks · Short answerBoard level

From a point $P$ , the length of the tangent to a circle with centre $O$ is $24$ cm. The nearest point of the circle to $P$ is $16$ cm from $P$ . Find the radius of the circle.

Try it first, then show the solutionSolution

Let radius be $r$ . The nearest point lies on $P O$ , so $O P = 16 + r$ . (½)
$∠ O T P = 9 0^{\circ}$ (tangent $⊥$ radius at $T$ ). (½)
$(16 + r)^{2} = r^{2} + 2 4^{2}$ (Pythagoras). (½)
$256 + 32 r + r^{2} = r^{2} + 576$ . (½)
$32 r = 320$ , so $r = 10$ cm. (½)
Check: $O P = 26$ , $1 0^{2} + 2 4^{2} = 676 = 2 6^{2}$ . (½)

Answer: Radius = 10 cm

Example 3: Concentric circles chord and tangent · 3 marks · Short answerHarder board level

Two concentric circles have centre $O$ and radii $13$ cm and $5$ cm. Chord $A B$ of the larger circle touches the smaller circle at $C$ . A point $D$ on the larger circle is such that $A D$ is a diameter of the larger circle. Find $A B$ and $B D$ .

Try it first, then show the solutionSolution

$O C ⊥ A B$ (tangent to smaller circle $⊥$ radius $O C$ ). (½)
Perpendicular from centre bisects chord $A B$ of the larger circle, so $A C = C B$ . (½)
$A C = 1 3^{2} - 5^{2} = 12$ cm, so $A B = 24$ cm. (½)
In $△ A B D$ , $O$ and $C$ are midpoints of $A D$ and $A B$ . (½)
By the midpoint theorem, $B D = 2 O C = 10$ cm. (½)
Check: $∠ A B D = 9 0^{\circ}$ (angle in semicircle), $2 4^{2} + 1 0^{2} = 676 = 2 6^{2}$ . (½)

Answer: AB = 24 cm and BD = 10 cm

Example 4: Tangents, chord of contact and an angle · 5 marks · Long answerHarder board level

(i) Prove that the lengths of tangents drawn from an external point to a circle are equal.
(ii) $P A$ and $P B$ are tangents from $P$ to a circle with centre $O$ and radius $6$ cm, with $∠ A P B = 6 0^{\circ}$ . Find $O P$ , $P A$ and the length of the chord $A B$ .

Try it first, then show the solutionSolution

(i) Given: tangents $P A$ , $P B$ from $P$ to circle with centre $O$ . To prove: $P A = P B$ . Construction: join $O A$ , $O B$ , $O P$ . (½)
In $△ O A P$ and $△ O B P$ : $∠ O A P = ∠ O B P = 9 0^{\circ}$ (tangent $⊥$ radius), $O A = O B$ (radii), $O P$ common. (1)
So $△ O A P ≅ △ O B P$ (RHS), hence $P A = P B$ (CPCT). (½)
(ii) $O P$ bisects $∠ A P B$ (from the congruence), so $∠ O P A = 3 0^{\circ}$ . (½)
$sin 3 0^{\circ} = \frac{O A}{O P}$ gives $O P = 12$ cm. (½)
$P A = 144 - 36 = 63$ cm. (½)
$P A = P B$ and $∠ A P B = 6 0^{\circ}$ , so $△ P A B$ is equilateral. (1)
$A B = P A = 63$ cm. (½)

Answer: OP = 12 cm, PA = 6√3 cm, AB = 6√3 cm

Example 5: Case study: belt around a wheel · 4 marks · Case-basedHOTS / competency

Imagine a machine in which a belt leaves a circular wheel with centre $O$ along two straight parts $P A$ and $P B$ that meet at a hook $P$ . The straight parts touch the wheel at $A$ and $B$ . Each straight part is $48$ cm long, and the hook is $32$ cm from the nearest point of the wheel's rim. A straight metal guard $C D$ is fixed so that it touches the wheel at a point $E$ on the minor arc $A B$ , with $C$ on $P A$ and $D$ on $P B$ .
(i) Why is $∠ O A P = 9 0^{\circ}$ ? (1)
(ii) Find the radius of the wheel. (1)
(iii) Find the perimeter of $△ P C D$ . (2)

Try it first, then show the solutionSolution

(i) $P A$ is a tangent at $A$ and the tangent at any point is perpendicular to the radius through the point of contact. (1)
(ii) Let the radius be $r$ . The nearest point of the rim lies on $P O$ , so $O P = r + 32$ , and $(r + 32)^{2} = r^{2} + 4 8^{2}$ (Pythagoras, right angle at $A$ ). (½)
$64 r + 1024 = 2304$ , so $64 r = 1280$ and $r = 20$ cm. Check: $2 0^{2} + 4 8^{2} = 2704 = 5 2^{2}$ . (½)
(iii) $C A = C E$ (tangents from $C$ ) and $D B = D E$ (tangents from $D$ ). (½)
Perimeter $= P C + C E + E D + D P = P C + C A + D B + D P$ . (½)
$= P A + P B = 48 + 48$ (since $P C + C A = P A$ and $D P + D B = P B$ ). (½)
$= 96$ cm. (½)

Answer: (i) tangent ⊥ radius at the point of contact; (ii) 20 cm; (iii) 96 cm

Example 6: Incircle of a right triangle · 3 marks · Short answerHarder board level

In $△ A B C$ , $∠ B = 9 0^{\circ}$ , $A B = 8$ cm and $B C = 6$ cm. A circle with centre $O$ is drawn inside the triangle touching $A B$ , $B C$ and $C A$ at $P$ , $Q$ and $R$ . Find the radius of the circle and the lengths $A P$ , $B Q$ and $C R$ .

Try it first, then show the solutionSolution

$A C = 8^{2} + 6^{2} = 10$ cm (Pythagoras). (½)
$∠ O P B = ∠ O QB = 9 0^{\circ}$ (tangent $⊥$ radius) and $∠ B = 9 0^{\circ}$ , with $O P = O Q = r$ , so $O P B Q$ is a square and $B P = B Q = r$ . (½)
$A P = 8 - r$ and $A R = A P = 8 - r$ (tangents from $A$ ). (½)
$C Q = 6 - r$ and $C R = C Q = 6 - r$ (tangents from $C$ ). (½)
$A R + C R = A C$ : $(8 - r) + (6 - r) = 10$ , so $r = 2$ cm. (½)
$A P = 6$ cm, $B Q = 2$ cm, $C R = 4$ cm. Check: area $= \frac{1}{2} \cdot 8 \cdot 6 = 24$ and $r \times$ semi-perimeter $= 2 \times 12 = 24$ . (½)

Answer: r = 2 cm, AP = 6 cm, BQ = 2 cm, CR = 4 cm

Example 7: Circumscribed quadrilateral: proof and algebraic sides · 5 marks · Long answerHarder board level

(i) A circle touches all four sides $A B$ , $B C$ , $C D$ and $D A$ of a quadrilateral $A B C D$ . Prove that $A B + C D = A D + B C$ .
(ii) In such a quadrilateral, $A B = (2 y + 1)$ cm, $B C = (y + 9)$ cm, $C D = (3 y - 2)$ cm and $D A = (y + 5)$ cm. Find $y$ , the four sides and the perimeter.

Try it first, then show the solutionSolution

(i) Given: the circle touches $A B$ , $B C$ , $C D$ , $D A$ at $P$ , $Q$ , $R$ , $S$ . To prove: $A B + C D = A D + B C$ . (½)
$A P = A S$ , $B P = B Q$ , $C R = C Q$ , $D R = D S$ (tangents from an external point are equal). (1)
Adding: $(A P + B P) + (C R + D R) = (A S + D S) + (B Q + C Q)$ . (1)
So $A B + C D = A D + B C$ . (½)
(ii) $(2 y + 1) + (3 y - 2) = (y + 5) + (y + 9)$ , so $5 y - 1 = 2 y + 14$ . (½)
$3 y = 15$ , so $y = 5$ . (½)
$A B = 11$ cm, $B C = 14$ cm, $C D = 13$ cm, $D A = 10$ cm. Check: $11 + 13 = 24 = 10 + 14$ . (½)
Perimeter $= 11 + 14 + 13 + 10 = 48$ cm. (½)

Answer: y = 5; AB = 11 cm, BC = 14 cm, CD = 13 cm, DA = 10 cm; perimeter = 48 cm

Example 8: MCQ: angle between tangents from a chord angle · 1 mark · MCQBoard level

$P A$ and $P B$ are tangents from an external point $P$ to a circle with centre $O$ . If $∠ O A B = 2 5^{\circ}$ , then $∠ A P B$ is
(a) $2 5^{\circ}$
(b) $5 0^{\circ}$
(c) $6 5^{\circ}$
(d) $13 0^{\circ}$

Try it first, then show the solutionSolution

$O A = O B$ (radii), so $∠ O B A = 2 5^{\circ}$ and $∠ A O B = 18 0^{\circ} - 5 0^{\circ} = 13 0^{\circ}$ . (½)
$∠ A P B = 18 0^{\circ} - 13 0^{\circ} = 5 0^{\circ}$ (kite $O A P B$ has two right angles). (½)

Answer: (b) 50°

Example 9: Assertion-reason: equal tangents · 1 mark · Assertion-reasonBoard level

Assertion (A): $P A$ and $P B$ are tangents from $P$ to a circle, with $P A = 5$ cm and $∠ A P B = 6 0^{\circ}$ . Then the chord $A B = 5$ cm.
Reason (R): The lengths of tangents drawn from an external point to a circle are equal.
(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

$P A = P B$ , so $∠ P A B = ∠ P B A = \frac{18 0 ^{\circ} - 6 0 ^{\circ}}{2} = 6 0^{\circ}$ ; $△ P A B$ is equilateral and $A B = 5$ cm, so A is true. (½)
R is a true theorem and it is exactly the fact that makes $△ P A B$ isosceles, so R explains A. (½)

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

Where students lose marks in Solving Problems Using Tangent Properties

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

Mistake: Treating OP as a leg
Students subtract out of habit without locating the right angle.
Wrong working
$O P^{2} = P T^{2} - r^{2}$
Right working
$O P^{2} = P T^{2} + r^{2}$ since the right angle is at $T$
Do this: Mark $9 0^{\circ}$ at the contact point first; the side opposite it is $O P$ .
Usually costs: Usually all calculation marks in the part.
Mistake: Using distance to the circle as OP
The phrase 'distance from the circle' is read as 'distance from the centre'.
Wrong working
$P$ is 16 cm from the circle, so $O P = 16$
Right working
$O P = 16 + r$
Do this: Ask: is the distance measured to the centre or to the rim?
Usually costs: Wrong equation, so most of a 3-mark question.
Mistake: No reason for the right angle
The right angle looks obvious in the figure.
Wrong working
$∠ O A P = 9 0^{\circ}$ , so by Pythagoras...
Right working
$∠ O A P = 9 0^{\circ}$ (tangent $⊥$ radius through point of contact)
Do this: Write the reason in brackets every time.
Usually costs: ½ mark for the missing reason.
Mistake: Taking ∠APO equal to ∠APB
Forgetting that OP bisects the angle between tangents.
Wrong working
$tan 6 0^{\circ} = \frac{r}{P A}$ when $∠ A P B = 6 0^{\circ}$
Right working
$tan 3 0^{\circ} = \frac{r}{P A}$
Do this: Halve the angle between tangents before using trigonometry.
Usually costs: 1 to 2 marks.

Exam tips

Write Given, the reason for every right angle and every pair of equal tangents; step marks depend on them.
In MCQs on angles, use $∠ A P B + ∠ A O B = 18 0^{\circ}$ , then check with the isosceles triangle $P A B$ .
When a quadratic appears, reject the negative root in words, saying a length cannot be negative.
Check final lengths by substituting into Pythagoras; it takes ten seconds.
Construction of tangents is not in the syllabus, so no question asks you to draw them with compass steps.

Practice questions on Solving Problems Using Tangent Properties

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

Q1 · 3 marks · Short answerHarder board level
A quadrilateral $A B C D$ is drawn so that a circle touches all four of its sides. If $A B = (x + 3)$ cm, $B C = (x + 8)$ cm, $C D = (2 x - 1)$ cm and $D A = (x + 2)$ cm, find $x$ , each side and the perimeter of $A B C D$ .
Hint
Use AB + CD = AD + BC, which comes from the equal tangents at each vertex.
Answer
(x + 3) + (2x − 1) = (x + 2) + (x + 8), so 3x + 2 = 2x + 10 and x = 8. Sides AB = 11 cm, BC = 16 cm, CD = 15 cm, DA = 10 cm; check 11 + 15 = 26 = 10 + 16. Perimeter = 52 cm.
Q2 · 3 marks · Short answerBoard level
The tangent from $P$ to a circle of radius $r$ is $(r + 2)$ cm and $O P = (r + 4)$ cm. Find $r$ and the tangent length.
Hint
Set up (r + 4)² = r² + (r + 2)².
Answer
r² + 8r + 16 = 2r² + 4r + 4, so r² − 4r − 12 = 0, r = 6 cm (reject −2). Tangent = 8 cm, OP = 10 cm, check 36 + 64 = 100.
Q3 · 3 marks · Short answerHarder board level
A circle touches side $B C$ of $△ A B C$ at $D$ and touches $A B$ and $A C$ produced at $F$ and $E$ . If $A B = 9$ cm, $B C = 7$ cm and $C A = 8$ cm, find $A F$ and $B D$ .
Hint
AF = AE = half the perimeter, because BD = BF and CD = CE.
Answer
AF + AE = AB + BD + AC + CD = 9 + 8 + 7 = 24, so AF = 12 cm. BD = BF = AF − AB = 3 cm.
Q4 · 5 marks · Long answerHarder board level
Two circles with centres $O$ and $O^{'}$ and radii $8$ cm and $6$ cm intersect at $P$ and $Q$ , and $O P$ is a tangent to the second circle at $P$ .
(i) Find $O O^{'}$ .
(ii) Find the length of the common chord $P Q$ .
(iii) Show that $O Q$ is also a tangent to the second circle.
(iv) Find the area of the quadrilateral $O P O^{'} Q$ .
Hint
O'P ⊥ OP, so triangle OPO' is right-angled at P; OO' is the perpendicular bisector of PQ; compare triangles OPO' and OQO'.
Answer
(i) ∠OPO' = 90° (tangent ⊥ radius), so OO' = √(64 + 36) = 10 cm. (ii) OO' ⊥ PQ at M; area of triangle OPO' gives ½ × 8 × 6 = ½ × 10 × PM, so PM = 4.8 cm and PQ = 9.6 cm. (iii) OP = OQ = 8, O'P = O'Q = 6, OO' common, so triangles OPO' and OQO' are congruent (SSS) and ∠OQO' = 90°; OQ ⊥ radius O'Q at Q, so OQ is a tangent. (iv) Area = 2 × ½ × 8 × 6 = 48 cm²; check ½ × OO' × PQ = ½ × 10 × 9.6 = 48 cm².
Q5 · 4 marks · Short answerHOTS / competency
In $△ P QR$ , $∠ Q = 9 0^{\circ}$ . A circle with centre $O$ and radius $2$ cm lies inside the triangle and touches $P Q$ , $QR$ and $R P$ at $L$ , $M$ and $T$ . If $P T = 10$ cm, find $P Q$ , $QR$ and $P R$ , and the area of $△ P QR$ . Then verify that the area equals the radius multiplied by the semi-perimeter, and explain why.
Hint
OLQM is a square, so QL = QM = 2 cm. Let RM = RT = x and use Pythagoras.
Answer
∠OLQ = ∠OMQ = ∠Q = 90° and OL = OM, so OLQM is a square: QL = QM = 2 cm. PL = PT = 10 cm (tangents from P), so PQ = 12 cm. Let RM = RT = x; QR = x + 2, PR = x + 10. (x + 10)² = 12² + (x + 2)² gives 20x + 100 = 4x + 148, so x = 3. QR = 5 cm, PR = 13 cm. Area = ½ × 12 × 5 = 30 cm². Semi-perimeter = 15 cm and 2 × 15 = 30 cm², because the triangles OPQ, OQR and ORP each have height r on one side, so their areas add to ½ × r × perimeter.
Q6 · 1 mark · MCQBoard level
A point $P$ is $18$ cm from the nearest point of a circle of radius $7$ cm. The length of the tangent from $P$ to the circle is
(a) $17$ cm
(b) $24$ cm
(c) $25$ cm
(d) $26$ cm
Hint
First find the distance from P to the centre.
Answer
(b) 24 cm: OP = 18 + 7 = 25 cm, so the tangent = √(625 − 49) = √576 = 24 cm.
Q7 · 1 mark · MCQBoard level
A circle is inscribed in $△ A B C$ , touching $A B$ , $B C$ and $C A$ at $F$ , $D$ and $E$ . If $A F = 4$ cm, $B D = 6$ cm and $C E = 8$ cm, the perimeter of $△ A B C$ is
(a) $18$ cm
(b) $24$ cm
(c) $36$ cm
(d) $48$ cm
Hint
Each side is the sum of two tangent lengths from its ends.
Answer
(c) 36 cm: AE = AF = 4, BF = BD = 6, CD = CE = 8, so perimeter = 2(4 + 6 + 8) = 36 cm.

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 radius and tangent at the point of contact?Show answer
$9 0^{\circ}$
2.Tangent length formula?Show answer
$P T = O P^{2} - r^{2}$
3.If $∠ A P B = 8 0^{\circ}$ , what is $∠ A O B$ ?Show answer
$10 0^{\circ}$
4.Which line bisects the angle between two tangents?Show answer
The line joining the external point to the centre.
5.Chord of larger circle touching smaller concentric circle?Show answer
$2 R^{2} - r^{2}$

Solving Problems Using Tangent Properties: frequently asked questions

Which formula do I use most in circle numericals?

The tangent length relation from the right triangle formed by the radius, the tangent and the line from the centre to the external point. Combined with equal tangents, it solves most length questions.

Do I need trigonometry in this chapter?

Sometimes. If an angle between tangents is given, half of it sits in a right triangle with the radius, so sine or tangent ratios give the lengths.

Are constructions of tangents asked in 2026-27?

No. Constructions of tangents are not in the 2026-27 syllabus.

How much reasoning should I write in a 3-mark numerical?

Write each property you use in brackets, such as tangent perpendicular to radius or tangents from an external point. One short line per step with its reason is enough.