NCERT Solutions for Class 10 Maths Chapter 10 Ex 10.2 | Updated 2026-27

Key Concept: The tangent is perpendicular to the radius at the point of contact. So triangle OPQ is right-angled at P, where O is the centre, P is the point of contact, and Q is the external point.

Step 1: Identify the known values. Tangent length \( QP = 24 \) cm, distance from centre \( OQ = 25 \) cm, radius \( OP = r \) (unknown).

Step 2: Apply Pythagoras theorem in triangle OPQ:

\[ OQ^2 = OP^2 + QP^2 \]
\[ 25^2 = r^2 + 24^2 \]
\[ 625 = r^2 + 576 \]
\[ r^2 = 625 – 576 = 49 \]
\[ r = 7 \text{ cm} \]

Key Concept: Since TP and TQ are tangents, OP ⊥ TP and OQ ⊥ TQ. So \( \angle OPT = \angle OQT = 90° \).

Step 1: In quadrilateral OPTQ, the sum of all angles is 360°.

\[ \angle OPT + \angle PTQ + \angle TQO + \angle QOP = 360° \]
\[ 90° + \angle PTQ + 90° + 110° = 360° \]
\[ \angle PTQ + 290° = 360° \]
\[ \angle PTQ = 70° \]

Key Concept: OA ⊥ PA (radius ⊥ tangent), so \( \angle OAP = 90° \). Also, OP bisects \( \angle APB \) (by symmetry, since PA = PB).

Step 1: Since \( \angle APB = 80° \), and OP bisects this angle:

\[ \angle APO = \frac{80°}{2} = 40° \]

Step 2: In triangle OAP:

\[ \angle OAP + \angle APO + \angle POA = 180° \]
\[ 90° + 40° + \angle POA = 180° \]
\[ \angle POA = 50° \]

Given: AB is a diameter of a circle with centre O. PQ is the tangent at A and RS is the tangent at B.

To Prove: PQ ∥ RS

Step 1: Since PQ is a tangent at A and OA is the radius, by Theorem 1:

\[ OA \perp PQ \implies \angle OAP = 90° \text{ and } \angle OAQ = 90° \]

Step 2: Since RS is a tangent at B and OB is the radius:

\[ OB \perp RS \implies \angle OBR = 90° \text{ and } \angle OBS = 90° \]

Step 3: Since AB is a diameter, O lies on AB, so OA and OB are in the same straight line.

Step 4: \( \angle OAQ = \angle OBR = 90° \). These are alternate interior angles formed by the transversal AB with lines PQ and RS.

Why does this work? When a transversal cuts two lines and the alternate interior angles are equal (both 90°), the two lines are parallel.

Given: A tangent PQ to a circle at point A. A perpendicular drawn to PQ at A.

To Prove: This perpendicular passes through the centre O.

Step 1 (Proof by contradiction): Assume the perpendicular at A does NOT pass through the centre O. Let it pass through another point O’.

Step 2: Then O’A ⊥ PQ (by our assumption). But we also know that OA ⊥ PQ (radius is perpendicular to tangent at point of contact).

Step 3: This means both O’A and OA are perpendicular to PQ at the same point A. But through a given point, only one perpendicular can be drawn to a line.

Step 4: This is a contradiction. Therefore our assumption is wrong.

Why does this work? The uniqueness of the perpendicular from a point to a line ensures that O and O’ must be the same point.

Given: Distance from point A to centre O = 5 cm, tangent length AT = 4 cm, radius = OT = r.

Step 1: Since OT ⊥ AT, triangle OTA is right-angled at T. Apply Pythagoras theorem:

\[ OA^2 = OT^2 + AT^2 \]
\[ 5^2 = r^2 + 4^2 \]
\[ 25 = r^2 + 16 \]
\[ r^2 = 9 \]
\[ r = 3 \text{ cm} \]

Key Concept: A chord of the larger circle that is tangent to the smaller circle is bisected by the perpendicular from the common centre. The perpendicular from the centre to the chord equals the radius of the smaller circle.

Given: Radius of larger circle \( R = 5 \) cm, radius of smaller circle \( r = 3 \) cm. Let AB be the chord of the larger circle tangent to the smaller circle at point M.

Step 1: OM ⊥ AB (radius of smaller circle to point of tangency). So OM = 3 cm.

Step 2: In right triangle OMA, OA = 5 cm (radius of larger circle), OM = 3 cm:

\[ OA^2 = OM^2 + AM^2 \]
\[ 25 = 9 + AM^2 \]
\[ AM^2 = 16 \]
\[ AM = 4 \text{ cm} \]

Step 3: Since OM ⊥ AB, M is the midpoint of AB. Therefore:

\[ AB = 2 \times AM = 2 \times 4 = 8 \text{ cm} \]

Given: Quadrilateral ABCD circumscribes a circle. The circle touches sides AB, BC, CD, and DA at points P, Q, R, and S respectively.

To Prove: \( AB + CD = AD + BC \)

Step 1: Apply the theorem — tangents from an external point to a circle are equal in length.

Step 2: From vertex A: \( AP = AS \) (tangents from A)

From vertex B: \( BP = BQ \) (tangents from B)

From vertex C: \( CQ = CR \) (tangents from C)

From vertex D: \( DR = DS \) (tangents from D)

Step 3: Now add the left-hand side:

\[ AB + CD = (AP + PB) + (CR + RD) = (AS + BQ) + (CQ + DS) \]

Step 4: Rearrange the right-hand side:

\[ AD + BC = (AS + SD) + (BQ + QC) = (AS + DS) + (BQ + CQ) \]

Step 5: Both expressions equal \( AS + BQ + CQ + DS \). Therefore:

\[ AB + CD = AD + BC \]

Given: XY and X’Y’ are parallel tangents touching the circle at P and Q respectively. AB is another tangent touching at C, meeting XY at A and X’Y’ at B.

To Prove: \( \angle AOB = 90° \)

Step 1: From point A, two tangents are drawn: AP (along XY) and AC. By equal tangent theorem: \( AP = AC \). So triangle OAP ≅ triangle OAC (SSS or RHS). Therefore \( \angle POA = \angle COA \), i.e., OA bisects \( \angle POC \).

Step 2: From point B, two tangents are drawn: BQ (along X’Y’) and BC. By equal tangent theorem: \( BQ = BC \). So OB bisects \( \angle QOC \), i.e., \( \angle QOB = \angle COB \).

Step 3: Since XY ∥ X’Y’, PQ is a diameter (the line joining the two points of tangency of parallel tangents passes through the centre). Therefore \( \angle POQ = 180° \).

Step 4:

\[ \angle POC + \angle QOC = 180° \]
\[ 2\angle AOC + 2\angle BOC = 180° \]
\[ \angle AOC + \angle BOC = 90° \]
\[ \angle AOB = 90° \]

Given: PA and PB are tangents from external point P to a circle with centre O. A and B are the points of contact.

To Prove: \( \angle APB + \angle AOB = 180° \)

Step 1: Since OA ⊥ PA (radius ⊥ tangent): \( \angle OAP = 90° \)

Step 2: Since OB ⊥ PB (radius ⊥ tangent): \( \angle OBP = 90° \)

Step 3: In quadrilateral OAPB, the sum of all interior angles = 360°:

\[ \angle OAP + \angle APB + \angle PBO + \angle BOA = 360° \]
\[ 90° + \angle APB + 90° + \angle AOB = 360° \]
\[ \angle APB + \angle AOB = 180° \]

Given: ABCD is a parallelogram that circumscribes a circle. The circle touches AB at P, BC at Q, CD at R, and DA at S.

To Prove: ABCD is a rhombus, i.e., AB = BC = CD = DA.

Step 1: From Question 8’s result (proved for any circumscribed quadrilateral):

\[ AB + CD = BC + DA \quad \ldots (1) \]

Step 2: Since ABCD is a parallelogram, opposite sides are equal:

\[ AB = CD \quad \text{and} \quad BC = DA \quad \ldots (2) \]

Step 3: Substitute (2) into (1):

\[ AB + AB = BC + BC \]
\[ 2AB = 2BC \]
\[ AB = BC \]

Step 4: Since \( AB = CD \) (from (2)) and \( AB = BC \), and \( BC = DA \) (from (2)):

\[ AB = BC = CD = DA \]

Given: Circle of radius 4 cm inscribed in triangle ABC. BC is divided by point of contact D into BD = 8 cm and DC = 6 cm. Let the circle touch AB at E and AC at F.

Step 1: Using equal tangent theorem from each vertex:

From B: \( BE = BD = 8 \) cm

From C: \( CF = CD = 6 \) cm

From A: \( AE = AF = x \) cm (unknown)

Step 2: Express the sides in terms of x:

\[ AB = AE + BE = x + 8 \]
\[ AC = AF + FC = x + 6 \]
\[ BC = BD + DC = 8 + 6 = 14 \text{ cm} \]

Step 3: Calculate the semi-perimeter \( s \):

\[ s = \frac{AB + BC + CA}{2} = \frac{(x+8) + 14 + (x+6)}{2} = \frac{2x + 28}{2} = x + 14 \]

Step 4: The area of triangle ABC can be expressed two ways. Using the inscribed circle (inradius r = 4 cm):

\[ \text{Area} = r \times s = 4(x + 14) \]

Step 5: Also use Heron’s formula. With \( s = x + 14 \), \( s – a = s – BC = x \), \( s – b = s – AC = 8 \), \( s – c = s – AB = 6 \):

\[ \text{Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{(x+14)(x)(8)(6)} = \sqrt{48x(x+14)} \]

Step 6: Equate the two area expressions:

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

Square both sides:

\[ 16(x+14)^2 = 48x(x+14) \]
\[ 16(x+14) = 48x \]
\[ x + 14 = 3x \]
\[ 2x = 14 \]
\[ x = 7 \text{ cm} \]

Step 7: Find the sides:

\[ AB = x + 8 = 7 + 8 = 15 \text{ cm} \]
\[ AC = x + 6 = 7 + 6 = 13 \text{ cm} \]

Given: ABCD is a quadrilateral circumscribing a circle with centre O. The circle touches AB, BC, CD, and DA at P, Q, R, and S respectively.

To Prove: \( \angle AOB + \angle COD = 180° \) and \( \angle AOD + \angle BOC = 180° \)

Step 1: Join OP, OQ, OR, OS. In triangles OAP and OAS: OA = OA (common), AP = AS (equal tangents from A), OP = OS = r (radii). By SSS congruence: \( \triangle OAP \cong \triangle OAS \). Therefore \( \angle AOP = \angle AOS \).

Let \( \angle AOP = \angle AOS = \angle 1 \)

Step 2: Similarly, from vertex B: \( \angle BOP = \angle BOQ = \angle 2 \)

From vertex C: \( \angle COQ = \angle COR = \angle 3 \)

From vertex D: \( \angle DOR = \angle DOS = \angle 4 \)

Step 3: The angles around O sum to 360°:

\[ \angle 1 + \angle 2 + \angle 3 + \angle 4 + \angle 1 + \angle 2 + \angle 3 + \angle 4 = 360° \]
\[ 2(\angle 1 + \angle 2 + \angle 3 + \angle 4) = 360° \]
\[ \angle 1 + \angle 2 + \angle 3 + \angle 4 = 180° \]

Step 4: Now:

\[ \angle AOB = \angle 1 + \angle 2 \quad \text{and} \quad \angle COD = \angle 3 + \angle 4 \]
\[ \angle AOB + \angle COD = (\angle 1 + \angle 2) + (\angle 3 + \angle 4) = 180° \]

Step 5: Similarly:

\[ \angle BOC = \angle 2 + \angle 3 \quad \text{and} \quad \angle AOD = \angle 1 + \angle 4 \]
\[ \angle BOC + \angle AOD = 180° \]