CBSE Class 10 Mathematics Circles MCQs Set 09

Answer: (c) \( 8\sqrt{2} \) cm

Explanation:
1. Since the circles touch externally, OP = 8 + 4 = 12 cm. Join OQ and PR; both are perpendicular to the tangent QR.
2. Draw PS perpendicular to OQ, so QRPS is a rectangle. Then SP = QR and OS = 8 − 4 = 4 cm.
3. In the right triangle OSP, \( SP = \sqrt{12^2-4^2} = \sqrt{128} = 8\sqrt{2} \) cm. So QR = \( 8\sqrt{2} \) cm.

Teacher's Note:
1. A radius is perpendicular to the tangent at the point of contact.
2. Drawing the parallel from P to the tangent gives a right triangle to use Pythagoras.

Answer: (b) 32.5 cm

Explanation:
1. Let the radius be r and the centre O. The ball touches the ground at R, so OR is perpendicular to the ground and OR = r.
2. Taking the step corner P on the ball, OP = r. Horizontally P is 15 cm from OR and vertically it is (r − 10) cm below the level of O, so \( r^2 = 15^2 + (r-10)^2 \).
3. This gives 20r = 325, so r = 16.25 cm and the diameter is 32.5 cm.

Teacher's Note:
1. Use the right triangle formed by the centre, the corner P and the vertical through P.
2. The terms in r² cancel, leaving a simple equation.

Answer: (c) 30°

Explanation:
1. Since PQ is a tangent at Q, OQ is perpendicular to PQ, so ∠OQP = 90°.
2. ∠POQ and ∠POR form a linear pair, so ∠POQ = 180° − 120° = 60°.
3. In the triangle OPQ, ∠OPQ = 180° − 90° − 60° = 30°.

Teacher's Note:
1. The radius to the point of contact makes a right angle with the tangent.
2. Angles on a straight line add up to 180°.

Answer: (b) 6r

Explanation:
1. The six vertices divide the circle into six equal arcs, so each side subtends 60° at the centre.
2. Each triangle formed by the centre and one side has two sides equal to r and an angle of 60°, so it is equilateral and the side is r.
3. Perimeter = 6 × r = 6r.

Teacher's Note:
1. A regular hexagon has side equal to the radius of its circumscribed circle.
2. Perimeter = number of sides × side.

Answer: (b) 70°

Explanation:
1. The radii OP and OQ are perpendicular to the tangents, so ∠OPT = ∠OQT = 90°.
2. In the quadrilateral OPTQ, 110° + 90° + ∠PTQ + 90° = 360°.
3. So ∠PTQ = 70°.

Teacher's Note:
1. The angle between two tangents and the angle between the radii to the contact points are supplementary.
2. The angle sum of a quadrilateral is 360°.

Answer: (c) 60°

Explanation:
1. Let the tangents l (at A) and m (at B) meet at P. Tangents drawn from an external point are equal, so PA = PB.
2. Hence ∠PBA = ∠PAB = 60°, as the chord AB makes 60° with the tangent l.
3. In the triangle PAB, ∠APB = 180° − 60° − 60° = 60°.

Teacher's Note:
1. Equal tangents make the triangle PAB isosceles.
2. Equal sides have equal angles opposite to them.

Answer: (b) all chords are of the same length.

Explanation:
1. Let the radii be R and r, and let a chord of the outer circle touch the inner circle at a point M. The radius OM of the inner circle is perpendicular to the chord, so M is the mid-point of the chord.
2. \( \text{Half-chord} = \sqrt{R^2-r^2} \) by Pythagoras theorem in the right triangle formed by O, M and an end of the chord.
3. This value does not depend on which chord is chosen, so all such chords have the same length.

Teacher's Note:
1. The radius to a point of contact is perpendicular to the tangent, and a perpendicular from the centre bisects a chord.
2. The half-chord uses only R and r, which are fixed.

Answer: (b) 2

Explanation:
1. Two parallel tangents can be drawn to a circle in any given direction, one on each side of the circle (at the ends of the diameter perpendicular to that direction).
2. A secant fixes a direction, so there are exactly 2 tangents parallel to it.

Teacher's Note:
1. A circle has two tangents in each direction.
2. They touch the circle at the two ends of a diameter.

Answer: (b) 8 cm

Explanation:
1. From the external point D, the tangents DA and DC to the first circle are equal, so DA = DC = 4 cm.
2. Similarly the tangents DB and DC to the second circle are equal, so DB = DC = 4 cm.
3. AB = AD + DB = 4 + 4 = 8 cm.

Teacher's Note:
1. Tangents from one external point to a circle are equal in length.
2. Use the equality for each circle separately.

Answer: (d) 70°

Explanation:
1. Let the third tangent AB touch the circle at a point K. Tangents from A are equal, so OA bisects the angle between AB and the tangent line AT. Similarly OB bisects the angle between BA and BR.
2. The exterior angles of the triangle at A and B are 180° − ∠A and 180° − ∠B, and their sum is 180° + ∠P = 220°.
3. So ∠OAB + ∠OBA = 110° and ∠BOA = 180° − 110° = 70°. This equals 90° − ½ × 40° = 70°.

Teacher's Note:
1. The centre lies on the bisector of the angle between two tangents.
2. The result is ∠BOA = 90° − ∠P/2.

Answer: (c) 2.5

Explanation:
1. Circles touching externally have their centres at a distance equal to the sum of the radii, so the sides of the triangle are 1 + 2 = 3, 2 + 3 = 5 and 1 + 3 = 4.
2. Since 3² + 4² = 5², the triangle is right-angled and the hypotenuse is 5.
3. The circumradius of a right triangle is half the hypotenuse, so it is 2.5.

Teacher's Note:
1. Distance between centres of externally touching circles = sum of the radii.
2. A 3-4-5 triangle is right angled.

Answer: (a) 7 cm

Explanation:
1. The radius to the point of contact is perpendicular to the tangent, so the triangle formed by the centre, the point of contact and Q is right-angled at the point of contact.
2. \( r = \sqrt{25^2-24^2} = \sqrt{49} \) = 7 cm.

Teacher's Note:
1. Distance from the centre is the hypotenuse.
2. Tangent length² + radius² = distance².

Answer: (d) None of these

Explanation:
1. Statement (a): the tangent length is a leg of a right triangle whose hypotenuse is OP, so it is less than OP. It is correct.
2. Statement (b): the tangent at A is perpendicular to the radius OA. Since AB = AC, the line OA is the perpendicular bisector of BC, so BC is also perpendicular to OA and the two lines are parallel. It is correct.
3. Statement (c): the radii to the contact points and the tangents form a square of side a, so OP is its diagonal \( a\sqrt{2} \). It is correct.
4. Every statement is correct, so the answer is (d).

Teacher's Note:
1. Check each statement separately with a right triangle at the point of contact.
2. When all the statements are correct, 'None of these' is the answer to a 'not correct' question.

Answer: (d) None of these

Explanation:
1. Statement (a) is false: the angle between the tangents is 180° − 60° = 120°.
2. Statement (b) is false: if the point is at a distance 1.1r from the centre, the tangent length is r√(1.1² − 1) ≈ 0.46r, which is less than the radius.
3. Statement (c) is false: the centres lie on the perpendicular to PQ at A, which is not the perpendicular bisector of PQ unless A is the mid-point.
4. All three are false, so the answer is (d).

Teacher's Note:
1. Test each statement with a simple example.
2. The centres of circles touching a line at A lie on the perpendicular to the line at A.

Answer: (d) A line intersecting a circle in two points is called a chord.

Explanation:
1. Statements (a), (b) and (c) are standard properties of tangents: a tangent is perpendicular to the radius, a circle has only two tangents in a given direction, and these two touch at the ends of a diameter.
2. A line that cuts a circle in two points is a secant, not a chord. A chord is the line segment joining two points of the circle.
3. So the incorrect statement is (d).

Teacher's Note:
1. A secant is a line; a chord is a segment.
2. A tangent meets the circle at exactly one point.

Answer: (d) \( \sqrt{119} \) cm

Explanation:
1. OP is perpendicular to the tangent PQ, so the triangle OPQ is right-angled at P with OP = 5 cm and OQ = 12 cm.
2. \( PQ = \sqrt{OQ^2-OP^2} = \sqrt{144-25} = \sqrt{119} \) cm.

Teacher's Note:
1. The line through the centre is the hypotenuse.
2. Square both known sides before subtracting.

Answer: (b) 70°

Explanation:
1. OP ⟂ TP and OQ ⟂ TQ, so ∠OPT = ∠OQT = 90°.
2. In the quadrilateral TPOQ, ∠PTQ + 90° + 110° + 90° = 360°, so ∠PTQ = 70°.

Teacher's Note:
1. Opposite to the two right angles are the angle at T and the angle at O.
2. Those two angles are supplementary.

Answer: (a) 50°

Explanation:
1. ∠APB = 80° and the angle between the radii is ∠AOB = 180° − 80° = 100°.
2. Triangles OAP and OBP are congruent, so OP bisects ∠AOB and ∠POA = 100° ÷ 2 = 50°.

Teacher's Note:
1. The line joining the centre to the external point bisects the angle between the radii.
2. Check: in the right triangle OAP, ∠POA = 90° − 40° = 50°.

Answer: (b) 50°

Explanation:
1. The tangents at the ends of the two radii are perpendicular to those radii.
2. In the quadrilateral formed by the two radii and the two tangents, the angle between the tangents = 360° − 90° − 90° − 130° = 50°.

Teacher's Note:
1. Angle between tangents + angle between radii = 180°.
2. Use the angle sum of a quadrilateral.

Answer: (c) 5 cm

Explanation:
1. OP ⟂ AP and OQ ⟂ AQ, and ∠PAQ = 90°. Hence ∠POQ = 360° − 90° − 90° − 90° = 90°.
2. So APOQ has four right angles and AP = AQ, which makes it a square.
3. The radius OP equals the side AP = 5 cm.

Teacher's Note:
1. Three right angles in the quadrilateral force the fourth angle to be 90° as well.
2. Equal adjacent sides make the rectangle a square.