CBSE Class 10 Mathematics Circles MCQs Set 08

Answer: (b) Both the points lie on one of the tangents.

Explanation:
1. From each external point exactly 2 tangents can be drawn, so two external points normally give 4 tangents.
2. Only 3 tangents exist, so one tangent must be shared by both points. That means both points lie on the same tangent line.
3. So the answer is both the points lie on one of the tangents.

Teacher's Note:
1. Count 2 tangents per external point and ask when the count can drop to 3.
2. A shared tangent line is the only way to lose one tangent.

Answer: (a) 120°

Explanation:
1. Radius is perpendicular to the tangent at the point of contact, so ∠OAP = ∠OBP = 90°.
2. In quadrilateral OAPB the angles add up to 360°: ∠AOB = 360° − 90° − 90° − 60° = 120°.
3. So ∠AOB = 120°.

Teacher's Note:
1. Angle between two tangents and the angle at the centre are supplementary.
2. Here 180° − 60° = 120°.

Answer: (c) 8 cm

Explanation:
1. Let OP meet AB at M. M is the midpoint of AB, so AM = 4.8 cm and \( OM = \sqrt{6^2 - 4.8^2} = 3.6 \) cm.
2. In right \( \triangle OAP \) (right angle at A), AM is the altitude to OP, so \( OA^2 = OM \times OP \). Hence \( OP = 36 \div 3.6 = 10 \) cm.
3. \( PA = \sqrt{OP^2 - OA^2} = \sqrt{100 - 36} = 8 \) cm.

Teacher's Note:
1. The line joining the centre to the external point bisects the chord of contact at right angles.
2. Then use OA² = OM × OP in the right triangle OAP.

Answer: (d) infinite

Explanation:
1. A tangent at R is perpendicular to the radius through R, so the centre of any such circle lies on the line through R perpendicular to QR.
2. Every point of that line (on either side of R) is the centre of a circle that touches QR at R.
3. So infinitely many such circles can be drawn.

Teacher's Note:
1. A tangent line and its point of contact fix only the line on which the centre lies.
2. The radius is not fixed, so there are infinitely many circles.

Answer: (a) \( 60 \text{ cm}^2 \)

Explanation:
1. Tangent length \( PQ = \sqrt{13^2 - 5^2} = 12 \) cm.
2. Area of \( \triangle OQP = \tfrac{1}{2} \times 5 \times 12 = 30 \text{ cm}^2 \).
3. The quadrilateral PQOR is made of two such congruent triangles, so its area is 2 × 30 = 60 cm².

Teacher's Note:
1. Split the quadrilateral along OP into two congruent right triangles.
2. Do not forget to double the area of one triangle.

Answer: (b) 6 cm

Explanation:
1. The chord of the larger circle that touches the smaller circle is perpendicular to the radius of the smaller circle at the point of contact, so it is bisected there.
2. Half chord \( = \sqrt{5^2 - 4^2} = 3 \) cm.
3. The chord is 2 × 3 = 6 cm.

Teacher's Note:
1. Draw the radius to the point of contact: it bisects the chord.
2. Use the larger radius as the hypotenuse.

Answer: (c) \( 2\sqrt{3} \) cm

Explanation:
1. OA is perpendicular to the tangent AT, so △OAT is right-angled at A with hypotenuse OT = 4 cm.
2. \( AT = OT \cos 30^\circ = 4 \times \frac{\sqrt{3}}{2} = 2\sqrt{3} \) cm.

Teacher's Note:
1. Radius ⊥ tangent gives a right angle at A.
2. AT is the side adjacent to the 30° angle.

Answer: (d) 115°

Explanation:
1. OT ⊥ PT, so in △OTP: ∠TOP = 90° − 25° = 65°.
2. x is the angle on the straight line TO produced, so x = 180° − 65° = 115°.

Teacher's Note:
1. First find ∠TOP using the right angle at T.
2. Then use the linear pair on the diameter.

Answer: (c) 2

Explanation:
1. Two circles that intersect at two points have no tangent that separates them (no transverse common tangents) because each circle lies partly inside the other.
2. Only the two direct (outer) common tangents exist, so the number is 2.

Teacher's Note:
1. Recall: separate circles 4, touching externally 3, intersecting 2, touching internally 1.
2. Memorise this ladder.

Answer: (c) 11 cm

Explanation:
1. For a quadrilateral circumscribing a circle, the sums of opposite sides are equal: PQ + RS = QR + SP.
2. 12 + 14 = 15 + SP, so SP = 11 cm.

Teacher's Note:
1. Tangent lengths from each vertex are equal, which gives AB + CD = BC + DA.
2. Substitute the three known sides.

Answer: (b) 140°

Explanation:
1. The angle between a tangent and a chord equals the angle in the alternate segment, and the central angle on that chord is twice it.
2. So ∠MON = 2 × 70° = 140°.
3. Check: ∠OMN = 90° − 70° = 20°, ∠ONM = 20°, so ∠MON = 180° − 40° = 140°.

Teacher's Note:
1. Radius ⊥ tangent: ∠OMN = 90° − 70° = 20°.
2. Triangle OMN is isosceles, so find the apex angle.

Answer: (d) \( 3\sqrt{3} \) cm

Explanation:
1. The line OP from the centre to the external point bisects the angle between the tangents, so \( \angle OPA = 30^\circ \).
2. In right \( \triangle OAP \): \( \tan 30^\circ = \frac{OA}{PA} \), so \( PA = \frac{3}{1/\sqrt{3}} = 3\sqrt{3} \) cm.

Teacher's Note:
1. OP bisects the angle between the tangents.
2. Use tan of the half-angle in the right triangle OAP.

Answer: (c) \( \frac{9}{\sqrt{2}} \) cm

Explanation:
1. Let the radius of the circle with centre N be r. Then the radius of the other circle is 2r and \( ON = 2r + r = 3r \).
2. NX ⊥ OX, so in right △ONX: ON² = OX² + NX², i.e. 9r² = 324 + r².
3. \( 8r^2 = 324 \), so \( r^2 = \frac{81}{2} \) and \( r = \frac{9}{\sqrt{2}} \) cm.

Teacher's Note:
1. For circles touching externally, the distance between centres is the sum of radii.
2. The tangent from O to the small circle gives a right triangle ONX.

Answer: (d) 125°

Explanation:
1. OT ⊥ PT, so ∠TOP = 90° − 35° = 55°.
2. x lies on the straight line through T and O, so x = 180° − 55° = 125°.

Teacher's Note:
1. Find the angle at the centre first.
2. Then subtract from 180° for the angle on the straight line.

Answer: (a) \( \sqrt{2} \) cm

Explanation:
1. A, O, C lie on a straight line, so ∠AOB = 180° − 135° = 45°.
2. AB is a tangent, so \( \angle OAB = 90^\circ \) and \( \cos 45^\circ = \frac{OA}{OB} = \frac{x}{x^2} = \frac{1}{x} \).
3. So \( x = \sqrt{2} \): the radius is \( \sqrt{2} \) cm.

Teacher's Note:
1. Use the straight line AOC to get the angle at O.
2. Then use cos = adjacent / hypotenuse in the right triangle OAB.

Answer: (a) The radius of the smaller circle is half of the chord of the larger circle.

Explanation:
1. The chord touches the smaller circle, so it is a tangent there and is perpendicular to the radius through the point of contact (statements 2 and 3 are true).
2. The perpendicular from the centre to a chord bisects the chord, so statement 4 is true.
3. The half-chord is √(R² − r²), which is not r in general, so statement 1 cannot be said.

Teacher's Note:
1. Check each statement against tangent ⊥ radius and the perpendicular-bisects-chord property.
2. Statement 1 would need R² = 2r², which is not given.

Answer: (a) the perimeter of \( \triangle APQ \)

Explanation:
1. Tangents from one point are equal: PB = PR and QC = QR.
2. Perimeter of △APQ = AP + AQ + PQ = AP + AQ + PR + RQ = (AP + PB) + (AQ + QC) = AB + AC.
3. So AB + AC equals the perimeter of △APQ.

Teacher's Note:
1. Replace PR by PB and RQ by QC.
2. This idea is used again and again in tangent problems.

Answer: (b) 40°

Explanation:
1. The angle at the centre is twice the angle at the circumference on the same arc: ∠NOP = 2 × 25° = 50°.
2. ON ⊥ NQ, so ∠ONQ = 90°.
3. In △ONQ: ∠OQN = 180° − 90° − 50° = 40°.

Teacher's Note:
1. Use central angle = 2 × inscribed angle.
2. Then use the right angle between the radius and the tangent.

Answer: (c) 100°

Explanation:
1. Radius OP ⊥ tangent PR, so ∠OPQ = 90° − 50° = 40°.
2. OP = OQ, so ∠OQP = 40° and ∠POQ = 180° − 80° = 100°.

Teacher's Note:
1. Find the base angle of the isosceles triangle OPQ first.
2. The angle between a tangent and a chord is half the central angle on that chord.

Answer: (c) 60°

Explanation:
1. The line from the centre to the midpoint of a chord is perpendicular to the chord, so ∠APO = 90°.
2. In △AOP: ∠OAP = 90° − 60° = 30°, i.e. ∠BAC = 30°.
3. BC ⊥ AB (tangent ⊥ radius), so in △ABC: ∠C = 90° − 30° = 60°.

Teacher's Note:
1. Midpoint of a chord and the centre give a right angle.
2. Then use the right angle at B in triangle ABC.

Answer: (b) 21 cm

Explanation:
1. Join O to the points of contact; AB ⊥ AD and the radii are perpendicular to the sides, so APOS is a square and AP = AS = 10 cm.
2. CQ = CR = 27 cm (tangents from C), so BQ = 38 − 27 = 11 cm and BP = BQ = 11 cm.
3. x = AB = AP + PB = 10 + 11 = 21 cm.

Teacher's Note:
1. Opposite to the right angle at A the contact points form a square of side r.
2. Use equal tangent lengths from B and C.

Answer: (b) 50°

Explanation:
1. OA = OB (radii), so ∠OAB = ∠OBA = 40°.
2. OA ⊥ XY, so ∠OAY = 90° and ∠BAY = 90° − 40° = 50°.

Teacher's Note:
1. Find ∠OAB from the isosceles triangle OAB.
2. Subtract it from the right angle between the radius and the tangent.

Answer: (c) 61°

Explanation:
1. The angle at the circumference is half the angle at the centre: ∠ABQ = ½ × 58° = 29°.
2. AT ⊥ AB, so ∠BAT = 90°.
3. In △BAT: ∠ATB = 180° − 90° − 29° = 61°.

Teacher's Note:
1. Find ∠ABQ with the inscribed angle theorem.
2. Then use triangle BAT with the right angle at A.

Answer: (a) \( (15 - 5\sqrt{5}) \) cm

Explanation:
1. Take S = (0, 10), T = (20, 0), U = (20, 10). The circle touches SU and UT, so its centre is (20 − x, 10 − x) for radius x.
2. Line ST is \( x + 2y = 20 \). Distance of the centre from ST \( = \frac{20 - 3x}{\sqrt{5}} = x \), so \( 20 = x(3 + \sqrt{5}) \).
3. \( x = \frac{20}{3 + \sqrt{5}} = \frac{20(3 - \sqrt{5})}{4} = 15 - 5\sqrt{5} \approx 3.82 \) cm.

Teacher's Note:
1. Use coordinates: the circle lies in the corner at U.
2. Rationalise the denominator at the end.

Answer: (a) \( 4\sqrt{3} \) cm

Explanation:
1. The tangent at P is perpendicular to OP, so the parallel chord RS is perpendicular to OP; OQ bisects RS, so QS = 6 cm.
2. Let the radius be r: OQ = r/2. In right \( \triangle OQS \): \( r^2 = \left(\frac{r}{2}\right)^2 + 6^2 \).
3. \( \frac{3r^2}{4} = 36 \Rightarrow r^2 = 48 \Rightarrow r = 4\sqrt{3} \) cm.

Teacher's Note:
1. A line parallel to the tangent is perpendicular to the radius OP.
2. Apply Pythagoras in triangle OQS.

Answer: (c) 11 cm

Explanation:
1. DR = DS = 5 cm (tangents from D), so AR = 23 − 5 = 18 cm and AP = AR = 18 cm (tangents from A).
2. BP = AB − AP = 29 − 18 = 11 cm.
3. ∠B = 90° and the radii OP, OQ are perpendicular to the sides, so OPBQ is a square and r = BP = 11 cm.

Teacher's Note:
1. Equal tangents from A and D give AP.
2. With a right angle at B the centre, B and the two contact points form a square.

Answer: (c) 120°

Explanation:
1. OP ⊥ PT, so ∠OPQ = 90° − 60° = 30°; OP = OQ gives ∠OQP = 30°, so ∠POQ = 120°.
2. The major arc PQ subtends 240° at the centre, so the angle at R on the minor arc is half of 240° = 120°.
3. Check: ∠PRQ and the angle in the alternate segment (60°) are supplementary.

Teacher's Note:
1. Angles in opposite segments of a chord add up to 180°.
2. The tangent–chord angle equals the angle in the alternate segment.

Answer: (a) 60°, equilateral

Explanation:
1. In right \( \triangle OAP \): \( \sin \angle OPA = \frac{OA}{OP} = \frac{r}{2r} = \frac{1}{2} \), so \( \angle OPA = 30^\circ \).
2. OP bisects ∠APB, so ∠APB = 60°.
3. PA = PB (equal tangents) and the apex angle is 60°, so the other angles are 60° each: △APB is equilateral.

Teacher's Note:
1. Find the half-angle with sin = r/OP.
2. An isosceles triangle with an apex angle of 60° is equilateral.

Answer: (c) 20 cm

Explanation:
1. Tangents from one point are equal: CA = CE and DB = DE, and PA = PB = 10 cm.
2. Perimeter = PC + CD + PD = PC + CE + ED + PD = PC + CA + DB + PD = PA + PB = 10 + 10 = 20 cm.

Teacher's Note:
1. Break CD into CE + ED and replace by CA + DB.
2. The perimeter is always PA + PB, whatever the position of E.

Answer: (b) 6 cm

Explanation:
1. ∠OAP = 90° and ∠OAB = 30°, so ∠PAB = 60°; PA = PB gives ∠PBA = 60° too.
2. So △PAB is equilateral and PA = AB = 6 cm.

Teacher's Note:
1. Find the base angles of triangle PAB: 90° − 30° = 60°.
2. An equiangular triangle is equilateral.

Answer: (c) 8 cm

Explanation:
1. The chord touches the smaller circle, so it is perpendicular to the radius (3 cm) at the point of contact and is bisected there.
2. Half chord = √(5² − 3²) = 4 cm, so the chord = 8 cm.

Teacher's Note:
1. Draw the radius to the point of contact.
2. The larger radius is the hypotenuse.

Answer: (b) 10 cm

Explanation:
1. PR = PQ = 5 cm (tangents from P to the inner circle).
2. OQ ⊥ PS, so the perpendicular from the centre of the outer circle bisects the chord PS: PQ = QS = 5 cm.
3. PS = PQ + QS = 10 cm.

Teacher's Note:
1. The chord PS of the outer circle touches the inner circle at Q.
2. The point of contact is the midpoint of the chord.

Answer: (b) 13 cm

Explanation:
1. Let CQ = CR = x. Then AC = 6 + x and BC = 8 + x.
2. The perimeter of △ABC is 2(6 + 8 + x) = 28 + 2x, and area = ½ × r × perimeter, so 84 = ½ × 4 × (28 + 2x), giving x = 7.
3. AC = 6 + 7 = 13 cm (and BC = 15 cm).

Teacher's Note:
1. Use area = r × s, where s is the semi-perimeter.
2. The tangent lengths from A, B, C add up to s.

Answer: (c) 60°

Explanation:
1. OA ⊥ AT, so ∠AOT = 90° − 30° = 60°.
2. In △OAB, OA = OB and the apex angle ∠AOB = 60°, so ∠OBA = (180° − 60°) ÷ 2 = 60°.

Teacher's Note:
1. B lies on OT, so ∠AOB = ∠AOT.
2. An isosceles triangle with a 60° apex angle is equilateral.

Answer: (b) 16 cm

Explanation:
1. Tangents from one point are equal: XQ = XP = 10 cm and YR = YP = 6 cm.
2. QZ = XZ − XQ = 20 − 10 = 10 cm, so ZR = QZ = 10 cm.
3. YZ = YR + RZ = 6 + 10 = 16 cm.

Teacher's Note:
1. Mark equal tangent lengths on the figure.
2. Find the remaining tangent from Z using XZ.

Answer: (b) 2a

Explanation:
1. OP bisects the angle between the tangents, so ∠OPT = 30°, and OT = a is perpendicular to PT.
2. In right △OTP: sin 30° = OT/OP, so OP = a ÷ ½ = 2a.

Teacher's Note:
1. Use the half-angle 30° and sin = opposite / hypotenuse.
2. OP is the hypotenuse of the right triangle OTP.

Answer: (c) A is true, but R is false.

Explanation:
1. Tangents from one point are equal: AP = AR and BQ = BR, and XP = XQ.
2. XA + AR = XA + AP = XP and XB + BR = XB + BQ = XQ. Since XP = XQ, the Assertion is true.
3. No tangent can be drawn from a point inside a circle (every line through it cuts the circle at two points), so the Reason is false.
4. Hence A is true but R is false.

Teacher's Note:
1. Check each statement separately.
2. Tangents exist only from points outside (or on) the circle.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. The two parallel tangents are drawn at the two ends of a diameter; no third tangent can be parallel to them. So the Assertion is true.
2. A tangent is a line that meets the circle at exactly one point; this is its definition, so the Reason is true.
3. The definition does not explain why only two parallel tangents exist, so R is not the correct explanation of A.

Teacher's Note:
1. Both statements can be true without one explaining the other.
2. The number of parallel tangents comes from the ends of a diameter.

Answer: (c) A is true, but R is false.

Explanation:
1. OQ ⊥ PQ and OR ⊥ PR, so ∠QOR = 180° − ∠QPR.
2. In isosceles △OQR: ∠OQR = (180° − ∠QOR) ÷ 2 = ∠QPR ÷ 2, so ∠QPR = 2∠OQR. The Assertion is true.
3. ∠OQP is the angle between a radius and a tangent (90°); it does not subtend any arc, and ∠OPR is an angle of a different triangle. So the Reason is false.

Teacher's Note:
1. Write ∠QOR in terms of ∠QPR first.
2. Tangent–radius angles are right angles, not inscribed angles.

Answer: (d) A is false, but R is true.

Explanation:
1. Since QR is a diameter, the chords PQ and RS start at its two ends. The centre O is the midpoint of QR, and parallel chords from Q and R are at equal distances from O (on opposite sides).
2. Equal distances from the centre mean equal chords, so PQ is always equal to RS. The Assertion is false.
3. The Reason is a standard true theorem.
4. Hence A is false but R is true.

Teacher's Note:
1. Parallel chords through the two ends of a diameter are equidistant from the centre.
2. Chords at equal distances from the centre are equal.