NCERT Class 10 Maths – Exercise 10.2 Solutions

NCERT Class 10 Maths

Chapter 10 – Circles | Exercise 10.2

(Rationalized Syllabus 2025-26)

💡 Theorem 10.2

The lengths of tangents drawn from an external point to a circle are equal.
The tangent at any point of a circle is perpendicular to the radius through the point of contact.

From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is: (A) 7 cm (B) 12 cm (C) 15 cm (D) 24.5 cm

Let $O$ be the center and $T$ be the point of contact.

Given: $OQ = 25$ cm (Hypotenuse), Tangent $QT = 24$ cm.

Since radius $\perp$ tangent, $\triangle OTQ$ is a right triangle.

$$OT^2 + QT^2 = OQ^2$$

$$OT^2 + 24^2 = 25^2 \Rightarrow OT^2 + 576 = 625$$

$$OT^2 = 49 \Rightarrow OT = 7 \text{ cm}$$

✔️ (A) 7 cm

In Fig., if TP and TQ are the two tangents to a circle with centre O so that $\angle POQ = 110^\circ$, then $\angle PTQ$ is equal to: (A) $60^\circ$ (B) $70^\circ$ (C) $80^\circ$ (D) $90^\circ$

Tangents are perpendicular to radii: $\angle OPT = 90^\circ$ and $\angle OQT = 90^\circ$.

In quadrilateral $POQT$, the sum of angles is $360^\circ$.

$$\angle PTQ + \angle POQ + \angle OPT + \angle OQT = 360^\circ$$

$$\angle PTQ + 110^\circ + 90^\circ + 90^\circ = 360^\circ$$

$$\angle PTQ + 290^\circ = 360^\circ \Rightarrow \angle PTQ = 70^\circ$$

✔️ (B) $70^\circ$

If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of $80^\circ$, then $\angle POA$ is equal to: (A) $50^\circ$ (B) $60^\circ$ (C) $70^\circ$ (D) $80^\circ$

Given: $\angle APB = 80^\circ$.

Triangles $\triangle POA$ and $\triangle POB$ are congruent (RHS congruence).

Therefore, $OP$ bisects $\angle APB$. So, $\angle APO = 40^\circ$.

In $\triangle PAO$ ($\angle OAP = 90^\circ$):

$$\angle POA + \angle APO + \angle OAP = 180^\circ$$

$$\angle POA + 40^\circ + 90^\circ = 180^\circ$$

$$\angle POA = 180^\circ – 130^\circ = 50^\circ$$

✔️ (A) $50^\circ$

Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

Let $AB$ be the diameter of the circle with center $O$.

Let $PQ$ be the tangent at $A$ and $RS$ be the tangent at $B$.

Since radius $\perp$ tangent:

$$OA \perp PQ \Rightarrow \angle OAP = 90^\circ$$

$$OB \perp RS \Rightarrow \angle OBS = 90^\circ$$

Since $AB$ is a straight line, $\angle OAP$ and $\angle OBS$ are alternate interior angles.

As alternate interior angles are equal ($90^\circ$), the lines $PQ$ and $RS$ are parallel.

✔️ Proved

Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.

Proof by Contradiction:

Let tangent $AB$ touch the circle at $P$. Let us assume the perpendicular at $P$ does not pass through center $O$, but passes through another point $O’$.

So, $\angle O’PB = 90^\circ$ (by assumption).

But we know theorem 10.1: Radius is perpendicular to tangent. So $\angle OPB = 90^\circ$.

Comparing the two: $\angle O’PB = \angle OPB$.

This is only possible if line segments $O’P$ and $OP$ coincide.

Therefore, the perpendicular must pass through the center $O$.

✔️ Proved

The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.

Let $O$ be center, $T$ be point of contact. $OA = 5$ cm, $AT = 4$ cm.

In right $\triangle OTA$:

$$OT^2 + AT^2 = OA^2$$

$$r^2 + 4^2 = 5^2 \Rightarrow r^2 + 16 = 25$$

$$r^2 = 9 \Rightarrow r = 3 \text{ cm}$$

✔️ Radius is 3 cm.

Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.

Let $O$ be the common center. Let chord $AB$ of larger circle touch smaller circle at $P$.

Radius of smaller circle $OP = 3$ cm. Radius of larger circle $OA = 5$ cm.

Since tangent is $\perp$ radius, $OP \perp AB$. $OP$ bisects chord $AB$ (perpendicular from center bisects chord).

In right $\triangle OPA$:

$$AP^2 + OP^2 = OA^2$$

$$AP^2 + 3^2 = 5^2 \Rightarrow AP^2 = 25 – 9 = 16 \Rightarrow AP = 4$$

Total length $AB = 2 \times AP = 8$ cm.

✔️ Length of chord is 8 cm.

A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC.

Let the circle touch the sides $AB, BC, CD, DA$ at points $P, Q, R, S$ respectively.

Using the property that tangent lengths from an external point are equal:

$$AP = AS \quad \dots(1)$$

$$BP = BQ \quad \dots(2)$$

$$CR = CQ \quad \dots(3)$$

$$DR = DS \quad \dots(4)$$

Adding all equations:

$$(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)$$

$$AB + CD = AD + BC$$

✔️ Proved

In Fig., XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X’Y’ at B. Prove that $\angle AOB = 90^\circ$.

Join $O$ to point of contact $C$.

In $\triangle OPA$ and $\triangle OCA$:

1. $OP = OC$ (Radii)

2. $OA = OA$ (Common)

3. $AP = AC$ (Tangents from A)

So $\triangle OPA \cong \triangle OCA$. Therefore $\angle POA = \angle COA$.

Similarly, $\triangle OQB \cong \triangle OCB$, so $\angle QOB = \angle COB$.

Since $POQ$ is a diameter (line through center perpendicular to parallel tangents), it is a straight line ($180^\circ$).

$$\angle POA + \angle COA + \angle COB + \angle QOB = 180^\circ$$

$$2\angle COA + 2\angle COB = 180^\circ$$

$$2(\angle COA + \angle COB) = 180^\circ \Rightarrow \angle AOB = 90^\circ$$

✔️ Proved

Q10

Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.

Let tangents be $PA$ and $PB$, center $O$. We need to prove $\angle APB + \angle AOB = 180^\circ$.

In quadrilateral $OAPB$:

$\angle OAP = 90^\circ$ and $\angle OBP = 90^\circ$ (Radius $\perp$ Tangent).

Sum of angles in quadrilateral is $360^\circ$:

$$\angle APB + \angle AOB + 90^\circ + 90^\circ = 360^\circ$$

$$\angle APB + \angle AOB = 180^\circ$$

✔️ Proved (Supplementary)

Q11

Prove that the parallelogram circumscribing a circle is a rhombus.

Let ABCD be a parallelogram circumscribing a circle.

From Q8, we know: $AB + CD = AD + BC$.

Since ABCD is a parallelogram, opposite sides are equal: $AB = CD$ and $AD = BC$.

Substituting these:

$$2AB = 2AD \Rightarrow AB = AD$$

Since adjacent sides are equal in a parallelogram, all sides are equal.

✔️ Therefore, ABCD is a rhombus.

Q12

A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively. Find the sides AB and AC.

Step 1: Setup

Let contact points on $AB$ be $F$ and $AC$ be $E$.

$CD = 6 \Rightarrow CE = 6$ cm.

$BD = 8 \Rightarrow BF = 8$ cm.

Let $AF = AE = x$ cm.

Sides: $a = 14$, $b = x+6$, $c = x+8$.

Step 2: Area by Heron’s Formula

$$s = \frac{14 + (x+6) + (x+8)}{2} = \frac{2x + 28}{2} = x + 14$$

$$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$$

$$= \sqrt{(x+14)(x)(8)(6)} = \sqrt{48x(x+14)}$$

Step 3: Area by Triangles ($\triangle OBC, \triangle OCA, \triangle OAB$)

Area = $\frac{1}{2}r(a+b+c) = \frac{1}{2} \times 4 \times (2x+28) = 2(2x+28) = 4(x+14)$.

Step 4: Equate and Solve

$$\sqrt{48x(x+14)} = 4(x+14)$$

Square both sides:

$$48x(x+14) = 16(x+14)^2$$

$$3x = x + 14 \quad (\text{dividing by } 16(x+14))$$

$$2x = 14 \Rightarrow x = 7$$

Step 5: Final Sides

$AB = x + 8 = 7 + 8 = 15$ cm.

$AC = x + 6 = 7 + 6 = 13$ cm.

✔️ AB = 15 cm, AC = 13 cm

Q13

Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Let quadrilateral be $ABCD$ touching circle at $P, Q, R, S$.

Join $O$ to all vertices and points of contact.

Using congruency (like $\triangle OAP \cong \triangle OAS$), adjacent angles at center are equal.

Let angles be $a, a, b, b, c, c, d, d$ in order around the center.

$$2(a+b+c+d) = 360^\circ \Rightarrow a+b+c+d = 180^\circ$$

Angle subtended by $AB = a+b$. Angle by $CD = c+d$.

Sum = $(a+b) + (c+d) = 180^\circ$.

✔️ Proved

🎉 Exercise 10.2 Completed | Chapter 10 Circles