CG Board Class 10 Maths Chapter 12 Solutions: Circle and Tangents (वृत्त एवं स्पर्श रेखाएँ) (2026-27)

(i)** Radius = 10 cm and distance of chord from center = 6 cm.

• (ii) Radius = 15 cm and distance of chord from center = 9 cm.
• (ii) Given (दिया है): Radius r = OA = 15 cm, Distance from center d = OM = 9 cm.

(i)** 8 cm and 3 cm

• (ii) 14 cm and 24 cm
• (ii) Given (दिया है): Length of chord AB = 14 cm ⇒ AM = 14/2 = 7 cm, Distance OM = 24 cm.

Step 1:** Diameter PQ = 10 cm ⇒ Radius r = OP = O M = 10/2 = 5 cm.

• Step 2: Center is O. Given PR = 2 cm, distance OR = OP – PR = 5 – 2 = 3 cm.
• Step 3: In right-angled triangle △ ORM:

OM² = OR² + MR² ⇒ 5² = 3² + MR²

25 = 9 + MR² ⇒ MR² = 16 ⇒ MR = 4 cm

• Step 4: Since PQ ⊥ MN, PQ bisects MN:

MN = 2 × MR = 2 × 4 = 8 cm

• Final Answer: 8 cm

Step 1:** Given chord AB = 18 cm ⇒ AM = 18/2 = 9 cm.

• Step 2: Let radius of the circle be r. Center is O, so OA = OQ = r.
• Step 3: Distance OM = OQ – MQ = r – 3.
• Step 4: In right triangle △ OMA:

OA² = OM² + AM² ⇒ r² = (r – 3)² + 9²

r² = r² – 6r + 9 + 81

6r = 90 ⇒ r = 15 cm

• Final Answer: 15 cm

A circle with center O where chords AB = AC.

• सिद्ध करना है: AO is the angle bisector of ∠ BAC.
• प्रमाण:

Join OB, OC and OA.

In △ OAB and △ OAC:

1. AB = AC (Given)
2. OB = OC (Radii of the same circle)
3. OA = OA (Common side)

By SSS Congruence Axiom: △ OAB ≅ △ OAC.

Therefore, ∠ OAB = ∠ OAC (by CPCT).

Hence, AO bisects ∠ BAC, which proves that center O lies on the angle bisector of ∠ BAC.

(इति सिद्धम् / Hence Proved)

Step 1:** Let chords be AB = 24 cm and CD = 10 cm.

Half length of chords: AM = 24/2 = 12 cm, CN = 10/2 = 5 cm.

• Step 2: Let distance OM = x. Then distance ON = 17 – x.
• Step 3: In right triangles △ OMA and △ ONC:

r² = OM² + AM² = x² + 12² = x² + 144

r² = ON² + CN² = (17 – x)² + 5² = 289 – 34x + x² + 25

• Step 4: Equate both expressions for r²:

x² + 144 = x² – 34x + 314

34x = 314 – 144 = 170 ⇒ x = 5 cm

• Step 5: Calculate Radius r and Diameter d:

r² = 5² + 144 = 25 + 144 = 169 ⇒ r = √(169) = 13 cm

Diameter = 2r = 2 × 13 = 26 cm

• Final Answer: 26 cm

Circle with center O, PO bisects ∠ APD.

• रचना: Draw perpendiculars OL ⊥ AB and OM ⊥ CD.
• प्रमाण:

In right triangles △ OPL and △ OPM:

1. ∠ OLP = ∠ OMP = 90°(By construction)
2. ∠ OPL = ∠ OPM (Given, PO bisects ∠ APD)
3. OP = OP (Common side)

By AAS Congruence Axiom: △ OPL ≅ △ OPM.

Thus, OL = OM (By CPCT).

Since chords equidistant from the center are equal in length (Theorem 6):

AB = CD

(इति सिद्धम् / Hence Proved)

Step 1:** Let OC be the line segment joining centers where OC = 10 cm. The perpendicular bisector intersects OC at its midpoint M.

OM = OC/2 = 10/2 = 5 cm

• Step 2: Radius of bigger circle OA = 13 cm.
• Step 3: In right triangle △ OMA:

OA² = OM² + AM² ⇒ 13² = 5² + AM²

169 = 25 + AM² ⇒ AM² = 144 ⇒ AM = 12 cm

• Step 4: Length of chord AB = 2 × AM = 2 × 12 = 24 cm.
• Final Answer: 24 cm

Two equal chords AB and CD in a circle with center at O meet at right angle on point E. If P and Q are mid-points of chords AB and CD, then prove that OPEQ is a square.

हल:

• प्रमाण:

1. Since P and Q are midpoints of chords AB and CD, OP ⊥ AB and OQ ⊥ CD (Theorem 4). Thus, ∠ OPE = 90°and ∠ OQE = 90°.
2. Given that chords meet at right angles at E, so ∠ PEQ = 90°.
3. In quadrilateral OPEQ, three angles are 90°, hence fourth angle ∠ POQ = 360°- (90°+ 90°+ 90°) = 90°. Thus, OPEQ is a rectangle.
4. Since chords AB = CD are equal, they are equidistant from center O (Theorem 6): OP = OQ.
5. A rectangle with adjacent equal sides (OP = OQ) is a square.

(इति सिद्धम् / Hence Proved)

## Part 4: Solutions for Exercise – 3 (प्रश्नावली – 3)

Formula:** By Theorem 7, angle subtended by an arc at center is twice the angle subtended at the remaining circumference.

∠ AOB = 2 × ∠ ACB = 2 × 20°= 40°* Final Answer: 40°

Formula:** ∠ POQ = 2 × ∠ PRQ.

∠ POQ = 2 × 40°= 80°* Final Answer: 80°

Formula:** ∠ ACB = 1/2 ∠ AOB.

∠ ACB = (70°)/(2) = 35°* Final Answer: 35°

Calculation:** Based on angle subtended in same segment or semicircle property:

x = 25°* Final Answer: 25°

Calculation:** Using property of cyclic quadrilateral / segment angle relations:

∠ ADC = 75°* Final Answer: 75°

Calculation:**

Angle = 120°* Final Answer: 120°

Calculation:**

∠ QPR = 80°* Final Answer: 80°

Calculation:**

Length = 10 cm

• Final Answer: 10 cm

Step 1:** Opposite angles of a cyclic quadrilateral are supplementary (Theorem 10).

∠ DAB + ∠ DCB = 180° ⇒ 80°+ ∠ DCB = 180° ⇒ ∠ DCB = 100°* Step 2: Similarly:

∠ ABC + ∠ ADC = 180° ⇒ 80°+ ∠ ADC = 180° ⇒ ∠ ADC = 100°* Final Answer: ∠ DCB = 100°, ∠ ABC = 80°, ∠ ADC = 100°

Step 1:** Using opposite angle sum = 180°:

∠ QRS = 104°* Step 2: Angle in same segment:

∠ QTS = 104°* Final Answer: ∠ QRS = 104°, ∠ QTS = 104°

Calculation:**

∠ BAC = 60°* Final Answer: 60°

Calculation:**

x = 62°* Final Answer: 62°

Prove that the exterior angle of a cyclic quadrilateral is equal to its interior opposite angle.

हल:

• प्रमाण:

Let ABCD be a cyclic quadrilateral with side AB produced to E.

1. ∠ ABC + ∠ CBE = 180°(Linear pair axiom)
2. ∠ ABC + ∠ ADC = 180°(Opposite angles of cyclic quadrilateral)

Equating both:

∠ ABC + ∠ CBE = ∠ ABC + ∠ ADC ⇒ ∠ CBE = ∠ ADC

(Exterior angle = Interior opposite angle. इति सिद्धम् / Hence Proved)

Calculation:**

∠ QPR = 30°, ∠ PRS = 20°* Final Answer: ∠ QPR = 30°, ∠ PRS = 20°

Step 1:** Given distance from center OP = 10 cm, tangent length PT = 8 cm.

• Step 2: Radius is perpendicular to tangent at point of contact (Theorem 11): ∠ OTP = 90°.
• Step 3: In right-angled triangle △ OTP, using Pythagoras theorem:

OP² = OT² + PT² ⇒ 10² = r² + 8²

100 = r² + 64 ⇒ r² = 100 – 64 = 36

r = √36 = 6 cm

• Final Answer: 6 cm

Step 1:** In quadrilateral APOQ, ∠ APO = ∠ AQO = 90°(Theorem 11).

• Step 2: Sum of angles in quadrilateral APOQ = 360°:

∠ PAQ + ∠ POQ = 180° ⇒ ∠ PAQ + 100°= 180° ⇒ ∠ PAQ = 80°* Step 3: Line AO bisects ∠ PAQ (Theorem 12):

∠ PAO = (∠ PAQ)/(2) = (80°)/(2) = 40°* Final Answer: 40°

Prove that the tangents drawn at the endpoints of a diameter of a circle are parallel to each other.

हल:

• प्रमाण:

Let AB be a diameter with center O. Let lines L₁ and L₂ be tangents at endpoints A and B respectively.

Since radius is perpendicular to tangent at point of contact (Theorem 11):

∠ OAL₁ = 90° and ∠ OBL₂ = 90°These represent alternate interior angles along transversal line AB:

∠ OAL₁ = ∠ OBL₂ = 90°Since alternate interior angles are equal, tangent lines L₁ and L₂ are parallel to each other.

(इति सिद्धम् / Hence Proved)

A circle touches all sides of quadrilateral ABCD. Prove that AB + CD = BC + DA.

हल:

• प्रमाण:

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

Lengths of tangents drawn from an external point to a circle are equal (Theorem 12):

1. AP = AS
2. BP = BQ
3. CR = CQ
4. DR = DS

Adding all four equations:

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

AB + CD = BC + DA

(इति सिद्धम् / Hence Proved)

Prove that the angle between two tangents drawn from the same external point is supplementary to the angle subtended at the center by the chord joining their contact points.

हल:

• प्रमाण:

Let tangents AP and AQ touch circle at P and Q from external point A. Center is O.

Radii OP ⊥ AP and OQ ⊥ AQ (Theorem 11):

∠ APO = 90°, ∠ AQO = 90°In quadrilateral APOQ, sum of interior angles is 360°:

∠ PAQ + ∠ POQ + ∠ APO + ∠ AQO = 360°∠ PAQ + ∠ POQ + 90°+ 90°= 360° ⇒ ∠ PAQ + ∠ POQ = 180°Thus, ∠ PAQ and ∠ POQ are supplementary.

(इति सिद्धम् / Hence Proved)

Prove that the intercept of a tangent between two parallel tangents to a circle subtends a right angle at the center.

हल:

• प्रमाण:

Let two parallel tangents touch circle at X and Y. A third tangent line intercepts them at A and B and touches circle at C.

By congruency of triangles △ OAC ≅ △ OAX and △ OBC ≅ △ OBY:

∠ AOC = ∠ AOX and ∠ BOC = ∠ BOY

Since XOY is a straight line diameter:

∠ AOX + ∠ AOC + ∠ BOC + ∠ BOY = 180°2∠ AOC + 2∠ BOC = 180° ⇒ 2(∠ AOC + ∠ BOC) = 180°∠ AOB = 90°(इति सिद्धम् / Hence Proved)