RD Sharma Solutions Chapter 10 Circles Exercise 10.2 Class 10 Maths
Chapter Name | RD Sharma Chapter 10 Circles |
Book Name | RD Sharma Mathematics for Class 10 |
Other Exercises |
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Related Study | NCERT Solutions for Class 10 Maths |
Exercise 10.2 Solutions
1. If PT is a tangent at T to a circle whose center is O and OP = 17 cm, OT = 8 cm. Find the length of tangent segment PT.
Solution
OT = radius = 8 cm
OP = 17 cm
PT = length of tangent = ? 
2. Find the length of a tangent drawn to a circle with radius 5cm, from a point 13 cm from the center of the circle.
Solution
Consider a circle with center O.
OP = radius = 5cm.
A tangent is drawn at point P, such that line through O intersects it at Q, OB = 13cm.
Length of tangent PQ = ? 
3. A point P is 26 cm away from O of circle and the length PT of the tangent drawn from P to the circle is 10 cm. Find the radius of the circle.
Solution
Given OP = 26 cm
PT = length of tangent = 10 cm
radius = OT = ? 
Solution
Let the two circles intersect at points X and Y.
XY is the common chord.
Suppose 'A' is a point on the common chord and AM and AN be the tangents drawn A to the circle.
XY is the common chord.
Suppose 'A' is a point on the common chord and AM and AN be the tangents drawn A to the circle.
We need to show that AM = AN.
"Let PT be a tangent to the circle from an external point P and a secant to the circle through P intersects the circle at points A and B, then PT2 = PA × PB"
Now AM is the tangent and AXY is a secant
∴ AM2 = AX × AY ...(i)
AN is a tangent and AXY is a secant
AN is a tangent and AXY is a secant
∴ AN2 = AX × AY ...(ii)
7. Prove that the intercept of a tangent between two parallel tangents to a circle subtends a right angle at center. 
We know that
The two tangents drawn from external point to the circle are equal in length.
From point P, PA = PB = 14 cm
From point C, CE = CA
From, point D, DB = ED
Perimeter = PC + PD + CA + DB
The at point of contact, the tangent is perpendicular to the radius is line from center to point on circle. Therefore, perpendicular to tangent will pass through center of circle.
Let the circles be represented by (i) & (ii) respectively
TQ, TP are tangents to (i)
TP, TR are tangents to (ii)
We know that
The tangents drawn from external point to the circle will be equal in length.
For circle (i), TQ = TP ...(i)
For circle (ii), TP = TR ...(ii)
From (i) & (ii) TQ = TR
20. In fig. there are two concentric circles with Centre O of radii 5 cm and 3 cm. From an external point P, tangents PA and PB are drawn to these circles if AP = 12 cm, find the tangent length of BP.
22. In figure PA and PB are tangents from an external point p to the circle with centre O. LN touches the circle at M. Prove that PL + LM = PN + MN 
Given
O is Centre of circle
PA and PB are tangents
We know that
The tangents drawn from external point to the circle are equal in length.
From point P, PA = PB
⇒ PL + AL = PN + NB ...(i)
From point L & N, AL = LM and MN = NB
24. In the fig two tangents AB and AC are drawn to a circle O such that ∠BAC = 120° . Prove that OA = 2AB. 
Given
BC is tangent to circle
OE bisects AP, AE = EP
Consider ΔAOP
From (i) & (ii), we have AM2 = AN2
∴ AM = AN
∴ AM = AN
5. If the quadrilateral sides touch the circle prove that sum of pair of opposite sides is equal to the sum of other pair.
Solution
Consider a quadrilateral ABCD touching circle with center O at points E, F, G and H as in figure.
We know that
The tangents drawn from same external points to the circle are equal in length.
1. Consider tangents from point A [AM ⊥ AE]
AH = AE ...(i)
2. From point B [EB & BF]
BF = EB ...(ii)
3. From point C [CF & GC]
FC = CG ...(iii)
4. From point D [DG & DH]
DH = DG ...(iv)
Adding (i), (ii), (iii), & (iv)
(AH + BF + FC + DH) = [(AC + CB) + (CG + DG)]
⇒ (AH + DH) + (BF + FC) = (AE + EB) + (CG + DG)
⇒ AD + BC = AB + DC [from fig.]
Sum of one pair of opposite sides is equal to other.
We know that
The tangents drawn from same external points to the circle are equal in length.
1. Consider tangents from point A [AM ⊥ AE]
AH = AE ...(i)
2. From point B [EB & BF]
BF = EB ...(ii)
3. From point C [CF & GC]
FC = CG ...(iii)
4. From point D [DG & DH]
DH = DG ...(iv)
Adding (i), (ii), (iii), & (iv)
(AH + BF + FC + DH) = [(AC + CB) + (CG + DG)]
⇒ (AH + DH) + (BF + FC) = (AE + EB) + (CG + DG)
⇒ AD + BC = AB + DC [from fig.]
Sum of one pair of opposite sides is equal to other.
6. If AB, AC, PQ are tangents in Fig. and AB = 5cm find the perimeter of ΔAPQ.
Solution
Solution
Consider circle with center 'O' and has two parallel tangents through A & B at ends of diameter.
8. In Fig below, PQ is tangent at point R of the circle with center O. If ∠ TRQ = 30° . Find. ∠PRS.
Solution
9. If PA and PB are tangents from an outside point P. Such that PA = 10 cm and ∠APB = 60°. Find the length of chord AB.
Solution
AP = 10 cm
∠APB = 60°
Represented in the figure
We know that
A line drawn from center to point from where external tangents are drawn divides or bisects the angle made by tangents at that point ∠APO = ∠OPB = 1/2 × 60° = 30°
The chord AB will be bisected perpendicularly
∴ AB = 2AM
In ΔAMP,
∠APB = 60°
Represented in the figure
We know that
A line drawn from center to point from where external tangents are drawn divides or bisects the angle made by tangents at that point ∠APO = ∠OPB = 1/2 × 60° = 30°
The chord AB will be bisected perpendicularly
∴ AB = 2AM
In ΔAMP,
10. From an external point P, tangents PA and PB are drawn to the circle with centre O. If CD is the tangent to the circle at point E and PA = 14 cm. Find the perimeter of ABCD.
Solution
PA = 14 cm
Perimeter of ΔPCD = PC + PD + CD = PC + PD + CE + ED
Perimeter of ΔPCD = PC + PD + CD = PC + PD + CE + ED
We know that
The two tangents drawn from external point to the circle are equal in length.
From point P, PA = PB = 14 cm
From point C, CE = CA
From, point D, DB = ED
Perimeter = PC + PD + CA + DB
= (PC + CA) + (PD + DB)
= PA + PB = 14 + 14 = 28 cm.
= PA + PB = 14 + 14 = 28 cm.
11. In the fig. ABC is right triangle right angled at B such that BC = 6cm and AB = 8cm. Find the radius of its in circle.
Solution
BC = 6cm AB = 8cm
As ABC is right angled triangle
AC = 10 cm
Consider BQOP ∠B = 90° ,
∠BPO = ∠OQB = 90° [At point of contact, radius is perpendicular to tangent]
All the angles = 90° & adjacent sides are equal
∴ BQOP is square BP = BQ = r
We know that
The tangents drawn from any external point are equal in length.
AP = AR = AB - PB = 8 - r
QC = RC = BC - BQ = 6 - r
AC = AR + RC ⇒ 10 = 8 - r + 6 - r
⇒ 10 = 14 - 2r
⇒ 2r = 4
⇒ Radius = 2 cm
As ABC is right angled triangle
AC = 10 cm
Consider BQOP ∠B = 90° ,
∠BPO = ∠OQB = 90° [At point of contact, radius is perpendicular to tangent]
All the angles = 90° & adjacent sides are equal
∴ BQOP is square BP = BQ = r
We know that
The tangents drawn from any external point are equal in length.
AP = AR = AB - PB = 8 - r
QC = RC = BC - BQ = 6 - r
AC = AR + RC ⇒ 10 = 8 - r + 6 - r
⇒ 10 = 14 - 2r
⇒ 2r = 4
⇒ Radius = 2 cm
12. From a point P, two tangents PA and PB are drawn to a circle with center O. If OP = diameter of the circle shows that ΔAPB is equilateral.
Solution
OP = 2r
Tangents drawn from external point to the circle are equal in length PA = PB
14. If ΔABC is isosceles with AB = AC and C (0, 2) is the in circle of the ΔABC touching BC at L, prove that L, bisects BC.
Tangents drawn from external point to the circle are equal in length PA = PB
13. Two tangent segments PA and PB are drawn to a circle with center O such that ∠APB = 120° . Prove that OP = 2AP.
Solution
14. If ΔABC is isosceles with AB = AC and C (0, 2) is the in circle of the ΔABC touching BC at L, prove that L, bisects BC.
Solution
Given ΔABC is isosceles AB = AC
We know that
The tangents from external point to circle are equal in length
From point A, AP = AQ
But AB = AC
We know that
The tangents from external point to circle are equal in length
From point A, AP = AQ
But AB = AC
⇒ AP + PB = AQ + QC
⇒ PB = PC ...(i)
From B, PB = BL ...(ii)
from C, CL = CQ ...(iii)
From (i), (ii) & (iii)
BL = CL
∴ L bisects BC.
⇒ PB = PC ...(i)
From B, PB = BL ...(ii)
from C, CL = CQ ...(iii)
From (i), (ii) & (iii)
BL = CL
∴ L bisects BC.
15. In fig. a circle touches all the four sides of quadrilateral ABCD with AB = 6 cm, BC = 7 cm, CD = 4 cm. Find AD.
Solution
We know that the tangents drawn from any external point to circle are equal in length.
16. Prove that the perpendicular at the point of contact to a circle passes through the center of the circle.
Solution
We know that
The at point of contact, the tangent is perpendicular to the radius is line from center to point on circle. Therefore, perpendicular to tangent will pass through center of circle.
17. In fig.. O is the center of the circle and BCD is tangent to it at C. Prove that ∠BAC + ∠ACD = 90°
SolutionGiven
O is center of circle
BCD is tangent.
O is center of circle
BCD is tangent.
18. Two circles touch externally at a point P from a point T on the tangent at P, tangents TQ and TR are drawn to the circles with points of contact Q and E respectively. Prove that TQ = TR.
Solution
Let the circles be represented by (i) & (ii) respectively
TQ, TP are tangents to (i)
TP, TR are tangents to (ii)
We know that
The tangents drawn from external point to the circle will be equal in length.
For circle (i), TQ = TP ...(i)
For circle (ii), TP = TR ...(ii)
From (i) & (ii) TQ = TR
19. In the fig. a circle is inscribed in a quadrilateral ABCD in which ∠B = 90° if AD = 23 cm, AB = 29 cm an DS = 5 cm, find the radius of the circle.
Solution
Given AD = 23 cm
AB = 29 cm
∠ B = 90°
DS = 5 cm
From fig in quadrilateral POQB
∠OPB = ∠OQB = 90° = ∠B = ∠POQ
and PO = OQ
∴ POQB is a square PB = BQ = r
We know that
Tangents drawn from external point to circle are equal in length.
We know that
Tangents drawn from external point to circle are equal in length.
We know that
Tangents drawn from external point to circle are equal in length.
From A, AR = AQ ...(i)
From B, PB = QB ...(ii)
From C, PC = CS ...(iii)
From D, DR = DS ...(iv)
AB = 29 cm
∠ B = 90°
DS = 5 cm
From fig in quadrilateral POQB
∠OPB = ∠OQB = 90° = ∠B = ∠POQ
and PO = OQ
∴ POQB is a square PB = BQ = r
We know that
Tangents drawn from external point to circle are equal in length.
We know that
Tangents drawn from external point to circle are equal in length.
We know that
Tangents drawn from external point to circle are equal in length.
From A, AR = AQ ...(i)
From B, PB = QB ...(ii)
From C, PC = CS ...(iii)
From D, DR = DS ...(iv)
20. In fig. there are two concentric circles with Centre O of radii 5 cm and 3 cm. From an external point P, tangents PA and PB are drawn to these circles if AP = 12 cm, find the tangent length of BP.
Solution
Given
21. In fig. AB is chord of length 16 cm of a circle of radius 10 cm. The tangents at A and B intersect at a point P. Find the length of PA.
Solution
Given length of chord AB = 16 cm.
Radius OB = OA = 10 cm.
Let line through Centre to point from where tangents are drawn be intersecting chord AB at M. we know that the line joining Centre to point from where tangents are drawn be intersecting chord AB at M. we know that the line joining Centre to point from where tangents are drawn bisects the chord joining the points on the circle where tangents intersects the circle.
Radius OB = OA = 10 cm.
Let line through Centre to point from where tangents are drawn be intersecting chord AB at M. we know that the line joining Centre to point from where tangents are drawn be intersecting chord AB at M. we know that the line joining Centre to point from where tangents are drawn bisects the chord joining the points on the circle where tangents intersects the circle.
Solution
Given
O is Centre of circle
PA and PB are tangents
We know that
The tangents drawn from external point to the circle are equal in length.
From point P, PA = PB
⇒ PL + AL = PN + NB ...(i)
From point L & N, AL = LM and MN = NB
Substitute in (i)
PL + LM = PN + MN
Hence proved.
PL + LM = PN + MN
Hence proved.
23. In the fig. PO ⊥ QO. The tangents to the circle at P and Q intersect at a point T. Prove that PQ and OT are light bisectors of each other.
Solution
Solution
25. In the fig. BC is a tangent to the circle with Centre O. OE bisects AP. Prove that ΔAEO ~ ΔABC.
Solution
Given
BC is tangent to circle
OE bisects AP, AE = EP
Consider ΔAOP
26. The lengths of three consecutive sides of a quadrilateral circumscribing a circle are 4 cm, 5 cm and 7 cm respectively. Determine the length of fourth side.
Solution
27. In fig. common tangents PQ and RS to two circles intersect at A. Prove that PQ = RS.
Solution
Consider
28. Equal circles with centers O and O' touch each other at X. OO' produced to meet a circle with centre O' at A. AC is tangent to the circle whose centre is a O' D is perpendicular to AC. Find the value of DO'/CO
Solution
29. In figure OQ : PQ = 3 : 4 and perimeter of ΔPDQ = 60 cm. determine PQ, QR and OP.
Solution