CBSE Class 10 Mathematics Circles Assignment Set 10

Very Short Answer Type Questions

Question 1. In the given figure, XPY is a tangent to a circle with centre O, QR is a diameter of the circle and P lies on the circle. If \( \angle QPY = 60^\circ \), find \( \angle PQR \).

Answer:
\( OP \perp XY \) (radius is perpendicular to the tangent), so \( \angle OPY = 90^\circ \) and \( \angle OPQ = 90^\circ - \angle QPY = 90^\circ - 60^\circ = 30^\circ \).
In \( \triangle OPQ \), \( OP = OQ \) (radii), so \( \angle OQP = \angle OPQ = 30^\circ \).
Since QR is a diameter, \( \angle PQR = \angle OQP \). Hence \( \angle PQR = \) \( 30^\circ \).

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 2. Two circles having radii 2 cm and 3 cm touch each other internally. Find the length of the longest chord of the bigger circle which is tangent to the smaller circle.
Answer:
Let the bigger circle have centre O (radius 3 cm) and the smaller circle centre O' (radius 2 cm). They touch internally, so \( OO' = 3 - 2 = 1 \) cm.
A chord of the bigger circle that touches the smaller circle at P is longest when it is nearest to O. This happens when it is perpendicular to OO' at the point P of the smaller circle on the side of O, where \( O'P = 2 \) cm and \( OP = O'P - OO' = 2 - 1 = 1 \) cm.
The chord QR is perpendicular to OP at P, so P is the mid-point of QR. In the right \( \triangle OPR \): \( PR = \sqrt{OR^2 - OP^2} = \sqrt{3^2 - 1^2} = 2\sqrt{2} \) cm.
\( QR = 2 \times 2\sqrt{2} = 4\sqrt{2} \) cm. The longest chord is \( 4\sqrt{2} \) cm (about 5.66 cm).

Teacher's Note:
1. A chord is longer the nearer it is to the centre.
2. The perpendicular from the centre to a chord bisects the chord.

Question 3. If two tangents inclined at an angle of \( 60^\circ \) are drawn to a circle of radius 3 cm, then find the length of each tangent.
Answer:
Let the tangents PA and PB touch the circle (centre O) at A and B, with \( \angle APB = 60^\circ \). OP bisects the angle: \( \angle OPA = 30^\circ \), and \( \angle OAP = 90^\circ \).
In the right \( \triangle OAP \): \( \tan 30^\circ = \frac{OA}{AP} \Rightarrow \frac{1}{\sqrt{3}} = \frac{3}{AP} \Rightarrow AP = 3\sqrt{3} \) cm.
The length of each tangent is \( 3\sqrt{3} \) cm (about 5.2 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. In the given figure, PQ and PT are tangents from the point P to the circle, touching it at Q and T. LM is another tangent, touching the circle at N, with L on PQ and M on PT. If \( PQ = 28 \) cm, find the perimeter of \( \triangle PLM \).

Answer:
Tangents from an external point are equal. From P: \( PT = PQ = 28 \) cm. From L: \( LN = LQ \). From M: \( MN = MT \).
Perimeter of \( \triangle PLM = PL + LM + PM = PL + (LN + NM) + PM = (PL + LQ) + (MT + PM) = PQ + PT \).
\( = 28 + 28 = \) 56 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 5. PQ is a tangent to a circle with centre O at the point P. If \( \triangle OPQ \) is an isosceles triangle, then find \( \angle OQP \).
Answer:
The radius OP is perpendicular to the tangent PQ, so \( \angle OPQ = 90^\circ \).
In the isosceles triangle OPQ the right angle at P cannot be one of the two equal base angles (two right angles would exceed \( 180^\circ \)), so the equal angles are \( \angle POQ \) and \( \angle OQP \).
\( \angle POQ = \angle OQP = \frac{180^\circ - 90^\circ}{2} = 45^\circ \). Hence \( \angle OQP = \) \( 45^\circ \).

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 6. If the radii of two concentric circles are 4 cm and 5 cm, then find the length of each chord of one circle which is tangent to the other circle.
Answer:
Let the chord AB of the larger circle touch the smaller circle at C. Then \( OC \perp AB \), \( OC = 4 \) cm and \( OA = 5 \) cm.
In the right \( \triangle OCA \): \( AC = \sqrt{5^2 - 4^2} = 3 \) cm. The perpendicular from the centre bisects the chord, so \( AB = 2 \times 3 = 6 \) cm.
The length of each such chord is 6 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.

Short Answer Type Questions

Question 7. Prove that the centre of a circle touching two intersecting lines lies on the angle bisector of the lines.
Answer:
Given: A circle with centre O touches two lines \( l_1 \) and \( l_2 \) intersecting at P, at M and N respectively. To prove: O lies on the bisector of the angle between \( l_1 \) and \( l_2 \).
Proof: Join OM, ON and OP. \( OM \perp l_1 \) and \( ON \perp l_2 \) (radius is perpendicular to the tangent), so \( \angle OMP = \angle ONP = 90^\circ \).
In \( \triangle OMP \) and \( \triangle ONP \): \( OM = ON \) (radii), \( OP = OP \) (common), \( \angle OMP = \angle ONP = 90^\circ \). So \( \triangle OMP \cong \triangle ONP \) (RHS) and \( \angle MPO = \angle NPO \) (C.P.C.T.).
Thus PO bisects \( \angle MPN \), i.e. O lies on the bisector of the angle between the two lines. 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 8. The incircle of \( \triangle PQR \) touches the sides QR, RP and PQ at A, B and C respectively. Prove that \( PC + QA + RB = PB + RA + QC = \frac{1}{2}(\text{perimeter of } \triangle PQR) \).
Answer:
Tangents drawn from an external point are equal: \( PC = PB \) ...(i), \( QA = QC \) ...(ii), \( RB = RA \) ...(iii).
Adding (i), (ii) and (iii): \( PC + QA + RB = PB + QC + RA = a \) (say).
Perimeter of \( \triangle PQR = (PC + QC) + (QA + RA) + (RB + PB) = (PC + QA + RB) + (QC + RA + PB) = a + a = 2a \).
Hence \( PC + QA + RB = PB + RA + QC = a = \frac{1}{2}(\text{perimeter of } \triangle PQR) \). 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 9. A circle touches the side QR of \( \triangle PQR \) at A and touches PQ and PR produced at B and C respectively. Show that \( PB = \frac{1}{2}(\text{perimeter of } \triangle PQR) \).
Answer:
Tangents from an external point are equal: \( PB = PC \) ...(i) (from P), \( QA = QB \) ...(ii) (from Q), \( RA = RC \) ...(iii) (from R).
Perimeter of \( \triangle PQR = PQ + QR + PR = PQ + (QA + AR) + PR \). Using (ii) and (iii): \( = PQ + QB + RC + PR = (PQ + QB) + (PR + RC) = PB + PC \) (B lies on PQ produced and C on PR produced).
Using (i), perimeter \( = 2PB \). Hence \( PB = \frac{1}{2}(\text{perimeter of } \triangle PQR) \). 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 10. Prove that the tangents drawn at the ends of a chord of a circle make equal angles with the chord.
Answer:
Given: A circle with centre O and a chord AB. PA and PB are the tangents at A and B meeting at P. To prove: \( \angle PAB = \angle PBA \). Join OA and OB.
Proof: In \( \triangle OAB \), \( OA = OB \) (radii), so \( \angle OAB = \angle OBA \) ...(i)
\( \angle OAP = \angle OBP = 90^\circ \) (radius is perpendicular to the tangent). So \( \angle OAB + \angle PAB = \angle OBA + \angle PBA \).
Using (i): \( \angle PAB = \angle PBA \). 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 11. In the figure, ABCD is a quadrilateral circumscribing a circle with centre O and radius 10 cm, with \( AD \perp AB \). The circle touches AB at P, BC at Q, CD at R and DA at S. If \( CR = 27 \) cm and \( BC = 38 \) cm, find \( x = AB \).
Answer:
Join OP and OS. \( OP \perp AB \), \( OS \perp AD \) (radius is perpendicular to the tangent) and \( \angle A = 90^\circ \), so \( \angle POS = 90^\circ \) and APOS is a square: \( AP = AS = 10 \) cm.
Tangents from C are equal: \( CQ = CR = 27 \) cm, so \( BQ = BC - CQ = 38 - 27 = 11 \) cm. Tangents from B: \( BP = BQ = 11 \) cm.
\( x = AB = AP + PB = 10 + 11 = \) 21 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. In the given figure, \( \triangle PQR \) has \( PQ = PR \) and its incircle touches PQ at A, QR at B and RP at C. Prove that \( QB = BR \).
Answer:
Tangents from an external point are equal: \( PA = PC \), \( QA = QB \), \( RB = RC \) ...(i)
Given \( PQ = PR \). Subtract PA from both sides: \( PQ - PA = PR - PA \Rightarrow QA = PR - PC = RC \) (using \( PA = PC \)).
So \( QB = QA = RC = RB \) (using (i)). Hence \( QB = BR \). 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 given figure, PA and PB are tangents to a circle from an external point P. CD is another tangent touching the circle at Q, with C on PA and D on PB. If \( PA = 12 \) cm and \( QC = QD = 3 \) cm, then find \( PC + PD \).
Answer:
Tangents from C: \( CA = CQ = 3 \) cm. So \( PC = PA - CA = 12 - 3 = 9 \) cm.
Tangents from D: \( DB = DQ = 3 \) cm. Tangents from P: \( PB = PA = 12 \) cm, so \( PD = PB - DB = 12 - 3 = 9 \) cm.
\( PC + PD = 9 + 9 = \) 18 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 14. In the figure, AB is a chord of length 8 cm of a circle of radius 5 cm. The tangents to the circle at A and B intersect at P. Find the length of AP.

Answer:
Let O be the centre and M the mid-point of AB, so \( AM = 4 \) cm and \( OM \perp AB \). \( OM = \sqrt{OA^2 - AM^2} = \sqrt{25 - 16} = 3 \) cm.
P lies on OM produced (OP is the perpendicular bisector of AB). Let \( PM = x \) and \( AP = y \). In the right \( \triangle AMP \): \( y^2 = x^2 + 16 \) ...(i)
\( OA \perp AP \), so in the right \( \triangle OAP \): \( OP^2 = OA^2 + AP^2 \Rightarrow (x + 3)^2 = 25 + y^2 \) ...(ii)
Substituting (i) in (ii): \( x^2 + 6x + 9 = 25 + x^2 + 16 \Rightarrow 6x = 32 \Rightarrow x = \frac{16}{3} \). Then \( y^2 = \frac{256}{9} + 16 = \frac{400}{9} \Rightarrow y = \frac{20}{3} \) cm.
The length of AP is \( \frac{20}{3} \) cm (about 6.67 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 15. A circle is inscribed in \( \triangle ABC \) having sides \( BC = 8 \) cm, \( AB = 10 \) cm and \( AC = 12 \) cm, touching BC at L, AC at M and AB at N. Find the lengths BL, CM and AN.
Answer:
Tangents from an external point are equal. Let \( BL = BN = x \). Then \( CL = CM = 8 - x \) and \( AM = AN = 12 - (8 - x) = 4 + x \).
\( AB = AN + NB \Rightarrow 10 = (4 + x) + x \Rightarrow x = 3 \).
\( BL = 3 \) cm, \( CM = 8 - 3 = 5 \) cm and \( AN = 4 + 3 = 7 \) 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.

Long Answer Type Questions

Question 16. PQRS is a quadrilateral such that \( \angle S = 90^\circ \). A circle C(O, r) touches the sides PQ, QR, RS and SP at A, B, C and D respectively. If \( QR = 38 \) cm, \( RS = 25 \) cm and \( QA = 27 \) cm, find r.
Answer:
Tangents from Q: \( QB = QA = 27 \) cm, so \( RB = QR - QB = 38 - 27 = 11 \) cm. Tangents from R: \( RC = RB = 11 \) cm, so \( SC = RS - RC = 25 - 11 = 14 \) cm.
\( OC \perp RS \), \( OD \perp SP \) and \( \angle S = 90^\circ \), so OCSD is a square with \( OC = OD = SC = 14 \) cm.
Hence r = 14 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. A triangle PQR is drawn to circumscribe a circle of radius 8 cm such that the segments QT and TR into which QR is divided by the point of contact T are of lengths 14 cm and 16 cm respectively. If the area of \( \triangle PQR \) is 336 \( \text{cm}^2 \), find the sides PQ and PR.

Answer:
Let the circle touch PQ at S and PR at U, and let \( PS = PU = x \) cm (tangents from P). \( QS = QT = 14 \) cm and \( RU = RT = 16 \) cm. So \( QR = 30 \) cm, \( PQ = x + 14 \), \( PR = x + 16 \).
Area of \( \triangle PQR \) = area of \( \triangle OQR \) + area of \( \triangle OPQ \) + area of \( \triangle OPR \), each having height 8 cm (the radius): \( 336 = \frac{1}{2} \times 8 \times [30 + (x + 14) + (x + 16)] = 4(60 + 2x) \).
\( 336 = 240 + 8x \Rightarrow x = 12 \).
\( PQ = 12 + 14 = \) 26 cm and \( PR = 12 + 16 = \) 28 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 18. Tangents QA and QB are drawn from an external point Q to a circle with centre O, such that \( \angle AQB = 30^\circ \). A chord BC is drawn parallel to the tangent QA. Find \( \angle BAC \).
Answer:
\( QA = QB \) (tangents from Q), so \( \angle QAB = \angle QBA = \frac{180^\circ - 30^\circ}{2} = 75^\circ \).
\( BC \parallel AQ \) and AB is a transversal: \( \angle CBA = \angle QAB = 75^\circ \) (alternate angles).
\( OA \perp QA \) and \( OB \perp QB \), so in the quadrilateral OAQB: \( \angle AOB = 360^\circ - 90^\circ - 90^\circ - 30^\circ = 150^\circ \). The angle at the circumference on the same arc is half the angle at the centre: \( \angle ACB = \frac{150^\circ}{2} = 75^\circ \).
In \( \triangle ABC \): \( \angle BAC = 180^\circ - 75^\circ - 75^\circ = \) \( 30^\circ \).

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 19. \( a, b \) and \( c \) are the sides of a right triangle, where c is the hypotenuse. A circle of radius r touches the sides of the triangle. Prove that \( r = \frac{a + b - c}{2} \).
Answer:
Let the right angle be at C, \( BC = a \), \( CA = b \), \( AB = c \). The circle with centre O touches BC at M, CA at N and AB at P.
\( OM \perp BC \), \( ON \perp CA \), \( \angle C = 90^\circ \) and \( OM = ON = r \), so OMCN is a square: \( CM = CN = r \).
Tangents from A: \( AN = AP \). From B: \( BM = BP \). \( AN = b - r \), \( BM = a - r \). So \( c = AB = AP + PB = (b - r) + (a - r) \).
\( c = a + b - 2r \Rightarrow r = \frac{a + b - c}{2} \). 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 20. In the given figure, O is the centre of the circle. Determine \( \angle APC \), if DA and DC are tangents and \( \angle ADC = 50^\circ \), where P is a point on the arc of the circle nearer to D.

Answer:
\( OA \perp DA \) and \( OC \perp DC \), so \( \angle OAD = \angle OCD = 90^\circ \).
In the quadrilateral OADC: \( \angle AOC = 360^\circ - 90^\circ - 90^\circ - 50^\circ = 130^\circ \). The reflex angle \( \angle AOC = 360^\circ - 130^\circ = 230^\circ \).
The angle subtended by the arc APC at the centre is the reflex angle \( 230^\circ \), and the angle at P on the circle is half of it: \( \angle APC = \frac{230^\circ}{2} = \) \( 115^\circ \).

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.