NCERT Solutions for Exercise 10.2 Class 10 Maths Chapter 10 - Circles

Circles Class 10 Chapter 10 Circles Exercise: 10.2

Q1 From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the center is 25 cm. The radius of the circle is

(A) 7 cm

(B) 12 cm

(C) 15 cm

(D) 24.5 cm

Answer:

The correct option is (A) = 7 cm
Given that,
The length of the tangent (QT) is 24 cm and the length of OQ is 25 cm.
Suppose the length of the radius OT be cm.
We know that is a right angle triangle. So, by using Pythagoras theorem-

OT = 7 cm

Q2 In Fig. 10.11, if TP and TQ are the two tangents to a circle with center O so that POQ = , then PTQ is equal to

(A)

(B)

(C)

(D)

Answer:

The correct option is (b)

In figure,
Since POQT is quadrilateral. Therefore the sum of the opposite angles are 180

Q3 If tangents PA and PB from a point P to a circle with center O are inclined to each other at an angle of , then POA is equal to

(A) 50°

(B) 60°

(C) 70°

(D) 80°

Answer:

The correct option is (A)

It is given that, tangent PA and PB from point P inclined at
In triangle OAP and OBP

OA =OB (radii of the circle)
PA = PB (tangents of the circle)

Therefore, by SAS congruence

By CPCT,
Now, OPA = 80/4 = 40

In PAO,


= 50

Q4 Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

Answer:

Let line and line are two tangents of a circle and AB is the diameter of the circle.
OA and OB are perpendicular to the tangents and respectively.
therefore,

|| { 1 & 2 are alternate angles}

Q5 Prove that the perpendicular at the point of contact to the tangent to a circle passes through the center.

Answer:

In the above figure, the line AXB is the tangent to a circle with center O. Here, OX is the perpendicular to the tangent AXB ( ) at point of contact X.
Therefore, we have,
BXO + YXB =

OXY is a collinear
OX is passing through the center of the circle.

Q6 The length of a tangent from a point A at distance 5 cm from the center of the circle is 4 cm. Find the radius of the circle.

Answer:

Given that,
the length of the tangent from the point A (AP) is 4 cm and the length of OA is 5 cm.
Since APO = 90 ⁰
Therefore, APO is a right-angle triangle. By using Pythagoras theorem;

Q7 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:

In the above figure, Pq is the chord to the larger circle, which is also tangent to a smaller circle at the point of contact R.
We have,
radius of the larger circle OP = OQ = 5 cm
radius of the small circle (OR) = 3 cm

OR PQ [since PQ is tangent to a smaller circle]

According to question,

In OPR and OQR
PRO = QRO {both }

OR = OR {common}
OP = OQ {both radii}

By RHS congruence OPR OQR
So, by CPCT
PR = RQ
Now, In OPR,
by using pythagoras theorem,

PR = 4 cm
Hence, PQ = 2.PR = 8 cm

Q8 A quadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.12). Prove that AB + CD = AD + BC

Answer:

To prove- AB + CD = AD + BC
Proof-
We have,
Since the length of the tangents drawn from an external point to a circle are equal
AP =AS .......(i)
BP = BQ.........(ii)
AS = AP...........(iii)
CR = CQ ...........(iv)

By adding all the equations, we get;

AP + BP +RD+ CR = AS +DS +BQ +CQ
(AP + BP) + (RD + CR) = (AS+DS)+(BQ + CQ)
AB + CD = AD + BC

Hence proved.

Q9 In Fig. 10.13, XY and X'Y'are two parallel tangents to a circle with center O and another tangent AB with a point of contact C intersecting XY at A and X'Y' at B. Prove that AOB = 90°.

Answer:

To prove- AOB =
Proof-
In AOP and AOC,
OA =OA [Common]
OP = OC [Both radii]
AP =AC [tangents from external point A]
Therefore by SSS congruence, AOP AOC
and by CPCT, PAO = OAC
..................(i)

Similarly, from OBC and OBQ, we get;
QBC = 2. OBC.............(ii)

Adding eq (1) and eq (2)

PAC + QBC = 180
2( OBC + OAC) = 180
( OBC + OAC) = 90

Now, in OAB,
Sum of interior angle is 180.
So, OBC + OAC + AOB = 180
AOB = 90
hence proved.

Q10 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 center.

Answer:

To prove -
Proof-
We have, PA and PB are two tangents, B and A are the point of contacts of the tangent to a circle. And , (since tangents and radius are perpendiculars)

According to question,
In quadrilateral PAOB,
OAP + APB + PBO + BOA =
90 + APB + 90 + BOA = 360

Hence proved .

Q11 Prove that the parallelogram circumscribing a circle is a rhombus.

Answer:

To prove - the parallelogram circumscribing a circle is a rhombus
Proof-
ABCD is a parallelogram that circumscribes a circle with center O.
P, Q, R, S are the points of contacts on sides AB, BC, CD, and DA respectively
AB = CD .and AD = BC...........(i)
It is known that tangents drawn from an external point are equal in length.
RD = DS ...........(ii)
RC = QC...........(iii)
BP = BQ...........(iv)
AP = AS .............(v)
By adding eq (ii) to eq (v) we get;
(RD + RC) + (BP + AP) = (DS + AS) + (BQ + QC)
CD + AB = AD + BC
2AB = 2AD [from equation (i)]
AB = AD
Now, AB = AD and AB = CD
AB = AD = CD = BC

Hence ABCD is a rhombus.

Q12 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 (see Fig. 10.14). Find the sides AB and AC.

Answer:

Consider the above figure. Assume center O touches the sides AB and AC of the triangle at point E and F respectively.

Let the length of AE is x.

Now in ,

(tangents on the circle from point C)

(tangents on the circle from point B)

(tangents on the circle from point A)

Now AB = AE + EB

Now

Area of triangle

Now the area of

Area of

Area of

Now Area of the = Area of + Area of + Area of

On squaring both the side, we get

Hence

AB = x + 8

=> AB = 7+8

=> AB = 15

AC = 6 + x

=> AC = 6 + 7

=> AC = 13

Answer- AB = 15 and AC = 13

Q13 Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the center of the circle.

Answer:

Given- ABCD is a quadrilateral circumscribing a circle. P, Q, R, S are the point of contact on sides AB, BC, CD, and DA respectively.

To prove-

Proof -
Join OP, OQ, OR and OS
In triangle DOS and DOR,
OD =OD [common]
OS = OR [radii of same circle]
DR = DS [length of tangents drawn from an external point are equal ]
By SSS congruency, DOS DOR,
and by CPCT, DOS = DOR
.............(i)

Similarily,
...............(2, 3, 4)



SImilarily,

Hence proved.