CBSE Class 10 Mathematics Circles Assignment Set 09

NCERT Exercise 10.1 and 10.2

Question 1. How many tangents can a circle have?
Answer:
A circle is made up of infinitely many points, each at the same distance from the centre. One tangent can be drawn at every point of the circle.
Hence a circle has infinitely many tangents.

Teacher's Note:
1. Draw a neat figure and mark the right angles formed by the radius and the tangent.
2. Use tangents from an external point are equal and the Pythagoras theorem where lengths are asked.

Question 2. Fill in the blanks: (i) A tangent to a circle intersects it in ________ point(s). (ii) A line intersecting a circle in two points is called a ________. (iii) A circle can have ________ parallel tangents at the most. (iv) The common point of a tangent to a circle and the circle is called ________.
Answer:
(i) one
(ii) secant
(iii) two (one at each end of a diameter)
(iv) point of contact

Teacher's Note:
1. Draw a neat figure and mark the right angles formed by the radius and the tangent.
2. Use tangents from an external point are equal and the Pythagoras theorem where lengths are asked.

Question 3. A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that \( OQ = 12 \) cm. Length PQ is: (a) 12 cm (b) 13 cm (c) 8.5 cm (d) \( \sqrt{119} \) cm.

Answer:
The tangent is perpendicular to the radius at the point of contact: \( OP \perp PQ \).
In the right \( \triangle OPQ \): \( PQ = \sqrt{OQ^2 - OP^2} = \sqrt{144 - 25} = \sqrt{119} \) cm.
The correct option is (d) \( \sqrt{119} \) cm.

Teacher's Note:
1. Draw a neat figure and mark the right angles formed by the radius and the tangent.
2. Use tangents from an external point are equal and the Pythagoras theorem where lengths are asked.

Question 4. Draw a circle and two lines parallel to a given line such that one is a tangent and the other a secant to the circle.
Answer:
1. Draw a circle with centre O and a line AB (the given line).
2. Draw a diameter perpendicular to the given line AB, meeting the circle at M and N. Through M draw a line CD perpendicular to this diameter, i.e. parallel to AB. By Theorem 2 (a line perpendicular to a radius at its end), CD is a tangent at M.
3. Through any point on the diameter between O and N (other than the end N) draw a line EF perpendicular to the diameter. EF is parallel to AB and cuts the circle in two points, so it is a secant.
Thus CD is the tangent and EF the secant, both parallel to the given line AB.

Teacher's Note:
1. A line perpendicular to a radius at its outer end is a tangent.
2. Any parallel line that comes closer to the centre than the radius will cut the circle twice.

Question 5. 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.

Answer:
Let T be the point of contact; \( OT \perp QT \). In the right \( \triangle QTO \): \( OQ^2 = QT^2 + OT^2 \Rightarrow 25^2 = 24^2 + r^2 \Rightarrow r^2 = 625 - 576 = 49 \Rightarrow r = 7 \) cm.
The correct option is (a) 7 cm.

Teacher's Note:
1. Draw a neat figure and mark the right angles formed by the radius and the tangent.
2. Use tangents from an external point are equal and the Pythagoras theorem where lengths are asked.

Question 6. In the figure, 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) \( 90^\circ \).

Answer:
\( \angle OPT = \angle OQT = 90^\circ \) (radius is perpendicular to the tangent).
In the quadrilateral TPOQ the angles add up to \( 360^\circ \): \( \angle PTQ + 90^\circ + 110^\circ + 90^\circ = 360^\circ \Rightarrow \angle PTQ = 70^\circ \).
The correct option is (b) \( 70^\circ \).

Teacher's Note:
1. Draw a neat figure and mark the right angles formed by the radius and the tangent.
2. Use tangents from an external point are equal and the Pythagoras theorem where lengths are asked.

Question 7. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of \( 80^\circ \), then \( \angle POA \) is equal to (a) \( 50^\circ \) (b) \( 60^\circ \) (c) \( 70^\circ \) (d) \( 80^\circ \).

Answer:
\( \triangle POA \cong \triangle POB \) (RHS: \( \angle PAO = \angle PBO = 90^\circ \), \( OA = OB \), OP common). So OP bisects \( \angle APB \): \( \angle APO = \frac{1}{2} \times 80^\circ = 40^\circ \).
In \( \triangle PAO \): \( \angle POA = 180^\circ - 90^\circ - 40^\circ = 50^\circ \).
The correct option is (a) \( 50^\circ \).

Teacher's Note:
1. Draw a neat figure and mark the right angles formed by the radius and the tangent.
2. Use tangents from an external point are equal and the Pythagoras theorem where lengths are asked.

Question 8. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Answer:
Given: MN is a diameter of a circle with centre O. AB and CD are the tangents at M and N respectively. To prove: \( AB \parallel CD \).
Proof: The tangent is perpendicular to the radius through the point of contact: \( OM \perp AB \) and \( ON \perp CD \), so \( \angle OMA = \angle OMB = \angle ONC = \angle OND = 90^\circ \).
MN is a transversal, and \( \angle OMB = \angle OND \) (\( = 90^\circ \)) are alternate interior angles. Alternate interior angles being equal, \( AB \parallel CD \). Hence proved.
In simple words: Both tangents are perpendicular to the same diameter MN, so they are parallel.

Teacher's Note:
1. Draw a neat figure and mark the right angles formed by the radius and the tangent.
2. Use tangents from an external point are equal and the Pythagoras theorem where lengths are asked.

Question 9. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Answer:
Given: A circle with centre O and a tangent PR touching the circle at A. To prove: the perpendicular to PR at A passes through O.
Proof (by contradiction): Draw \( AQ \perp PR \) at A, and suppose its line meets the circle at Q (the other point of the circle on that line) but does not pass through O. Join OA.
The radius OA is perpendicular to the tangent at the point of contact, so \( \angle OAR = 90^\circ \). Also \( \angle QAR = 90^\circ \) (given). So \( \angle OAR = \angle QAR \), which is possible only when O lies on the line AQ.
Hence the perpendicular at the point of contact to the tangent passes through the centre.
In simple words: Both OA and the perpendicular at A are perpendicular to the tangent at A, so they are the same line.

Teacher's Note:
1. Draw a neat figure and mark the right angles formed by the radius and the tangent.
2. Use tangents from an external point are equal and the Pythagoras theorem where lengths are asked.

Question 10. 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.

Answer:
Let AP be the tangent, P the point of contact, \( OA = 5 \) cm and \( AP = 4 \) cm. \( \angle OPA = 90^\circ \).
\( OP^2 = OA^2 - AP^2 = 25 - 16 = 9 \Rightarrow OP = 3 \) cm.
The radius of the circle is 3 cm.

Teacher's Note:
1. Draw a neat figure and mark the right angles formed by the radius and the tangent.
2. Use tangents from an external point are equal and the Pythagoras theorem where lengths are asked.

Question 11. 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.

Answer:
Let PQ be the chord of the larger circle touching the smaller circle at S. Then \( OS \perp PQ \), \( OS = 3 \) cm and \( OP = 5 \) cm.
In the right \( \triangle OSP \): \( SP = \sqrt{5^2 - 3^2} = \sqrt{16} = 4 \) cm.
The perpendicular from the centre to a chord bisects the chord, so \( PQ = 2 \times SP = 8 \) cm.
The length of the chord is 8 cm.

Teacher's Note:
1. Draw a neat figure and mark the right angles formed by the radius and the tangent.
2. Use tangents from an external point are equal and the Pythagoras theorem where lengths are asked.

Question 12. A quadrilateral ABCD is drawn to circumscribe a circle. Prove that \( AB + CD = AD + BC \).

Answer:
Given: ABCD circumscribes a circle with centre O and touches the sides AB, BC, CD, DA at P, Q, R, S. To prove: \( AB + CD = AD + BC \).
Proof: Tangents drawn from an external point to a circle are equal. So \( AP = AS \), \( BP = BQ \), \( CR = CQ \), \( DR = DS \).
Adding: \( AP + BP + CR + DR = AS + BQ + CQ + DS \Rightarrow (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ) \).
\( AB + CD = AD + BC \). Hence proved.

Teacher's Note:
1. Draw a neat figure and mark the right angles formed by the radius and the tangent.
2. Use tangents from an external point are equal and the Pythagoras theorem where lengths are asked.

Question 13. In the figure, 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 \( \angle AOB = 90^\circ \).

Answer:
Given: XY and X'Y' are parallel tangents touching the circle at P and Q; AB is a tangent at C meeting XY at A and X'Y' at B. To prove: \( \angle AOB = 90^\circ \). Join OC.
In \( \triangle OPA \) and \( \triangle OCA \): \( OP = OC \) (radii), \( AO = AO \) (common), \( AP = AC \) (tangents from A). So \( \triangle OPA \cong \triangle OCA \) (SSS) and \( \angle POA = \angle COA \) ...(i)
Similarly \( \triangle OQB \cong \triangle OCB \) and \( \angle QOB = \angle COB \) ...(ii)
Since XY and X'Y' are parallel tangents, PQ is a diameter and POQ is a straight line: \( \angle POA + \angle COA + \angle COB + \angle QOB = 180^\circ \).
Using (i) and (ii): \( 2\angle COA + 2\angle COB = 180^\circ \Rightarrow \angle COA + \angle COB = 90^\circ \), i.e. \( \angle AOB = 90^\circ \). Hence proved.

Teacher's Note:
1. Draw a neat figure and mark the right angles formed by the radius and the tangent.
2. Use tangents from an external point are equal and the Pythagoras theorem where lengths are asked.

Question 14. 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.

Answer:
Given: Tangents QN and QM are drawn from the external point Q to a circle with centre O, touching it at N and M. To prove: \( \angle MQN + \angle NOM = 180^\circ \).
Proof: \( OM \perp QM \) and \( ON \perp QN \) (radius is perpendicular to the tangent), so \( \angle OMQ = \angle ONQ = 90^\circ \).
In the quadrilateral OMQN the angles add up to \( 360^\circ \): \( \angle MQN + 90^\circ + \angle NOM + 90^\circ = 360^\circ \).
Hence \( \angle MQN + \angle NOM = 180^\circ \). Hence proved.

Teacher's Note:
1. Draw a neat figure and mark the right angles formed by the radius and the tangent.
2. Use tangents from an external point are equal and the Pythagoras theorem where lengths are asked.

Question 15. Prove that a parallelogram circumscribing a circle is a rhombus.
Answer:
Given: A parallelogram ABCD circumscribes a circle, touching AB, BC, CD, DA at P, Q, R, S. To prove: ABCD is a rhombus, i.e. \( AB = BC = CD = DA \).
Proof: In a parallelogram, opposite sides are equal: \( AB = CD \) and \( AD = BC \) ...(i)
Tangents from an external point are equal: \( AP = AS \), \( BP = BQ \), \( CR = CQ \), \( DR = DS \). Adding: \( AP + BP + CR + DR = AS + BQ + CQ + DS \Rightarrow AB + CD = AD + BC \).
Using (i): \( 2AB = 2AD \Rightarrow AB = AD \) ...(ii)
From (i) and (ii), \( AB = BC = CD = DA \). So ABCD is a rhombus. Hence proved.

Teacher's Note:
1. Draw a neat figure and mark the right angles formed by the radius and the tangent.
2. Use tangents from an external point are equal and the Pythagoras theorem where lengths are asked.

Question 16. 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.

Answer:
Let the circle touch AB at E and AC at F, and let \( AE = AF = a \) (tangents from A). Also \( BE = BD = 8 \) cm and \( CF = CD = 6 \) cm. So \( AB = 8 + a \), \( AC = 6 + a \) and \( BC = 14 \) cm.
Semi-perimeter \( s = \frac{(8 + a) + (6 + a) + 14}{2} = 14 + a \). By Heron's formula, area \( = \sqrt{s(s - 14)(s - 6 - a)(s - 8 - a)} = \sqrt{(14 + a)(a)(8)(6)} = \sqrt{48a(14 + a)} \).
The area is also the sum of the areas of \( \triangle OBC, \triangle OCA, \triangle OAB \), each with height 4 cm (the radius): \( \frac{1}{2} \times 4 \times (14 + 6 + a + 8 + a) = 2(28 + 2a) = 56 + 4a \).
Equating: \( \sqrt{48a(14 + a)} = 56 + 4a = 4(14 + a) \Rightarrow 48a = 16(14 + a) \Rightarrow 3a = 14 + a \Rightarrow a = 7 \).
\( AB = 8 + 7 = \) 15 cm and \( AC = 6 + 7 = \) 13 cm.

Teacher's Note:
1. Draw a neat figure and mark the right angles formed by the radius and the tangent.
2. Use tangents from an external point are equal and the Pythagoras theorem where lengths are asked.

Question 17. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Answer:
Given: PQRS circumscribes a circle with centre O and touches the sides at A (on PQ), B (on QR), C (on RS) and D (on SP). To prove: \( \angle POQ + \angle SOR = 180^\circ \) and \( \angle QOR + \angle SOP = 180^\circ \).
Proof: Join OA, OB, OC, OD and the vertices to O. Tangents from the same point are equal and OA = OD (radii), OP common, so \( \triangle OAP \cong \triangle ODP \) (SSS) and \( \angle AOP = \angle DOP \). Mark the eight angles at O (two at each vertex direction): call them \( \angle 1 = \angle 8, \angle 2 = \angle 3, \angle 4 = \angle 5, \angle 6 = \angle 7 \), where \( \angle 1, \angle 2 \) lie at P and Q on one side of OA and so on, going once round O.
The eight angles add up to \( 360^\circ \): \( 2(\angle 1 + \angle 2 + \angle 5 + \angle 6) = 360^\circ \Rightarrow (\angle 1 + \angle 2) + (\angle 5 + \angle 6) = 180^\circ \).
Here \( \angle 1 + \angle 2 = \angle POQ \) and \( \angle 5 + \angle 6 = \angle ROS \), so \( \angle POQ + \angle ROS = 180^\circ \). In the same way \( \angle QOR + \angle SOP = 180^\circ \). Hence proved.

Teacher's Note:
1. Draw a neat figure and mark the right angles formed by the radius and the tangent.
2. Use tangents from an external point are equal and the Pythagoras theorem where lengths are asked.