 Lesson: Circles
Exercise 10.1 (4)
Question: 1
How many tangents can a circle have?
Solution
A circle can have infinite number of tangents.
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 ______.
Solution
(i) One
(ii) Secant (iii) Two
(iv) Point of contact
Question: 3
A tangent PQ at a point of a circle of radius 5 cm
P
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. cm
119
Solution
(d)
Draw a circle with radius 5 cm as shown in the figure
below.
Fig. Exc_10.1_3 The line from the centre to the tangent is
perpendicular to the tangent.
Thus, .
OP PQ
In , by Pythagoras theorem,
OPQ
2 2 2
OQ OP PQ
2 2
2
12 5 PQ
2
PQ 144 25
PQ 119
Thus, the length of PQ is .
119
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.
Solution
Step 1: Draw a line .
l
Step 2: Set the compass to the desired radius of the
circle.
Step 3: Put the needle point of the compass at the
circle’s centre. Step 4: Turn the compass through and draw
360
the circle above the line.
Step 5: Draw a line parallel to the given line such
that it intersects the circle at two points. It is a
secant of the circle.
Step 6: Draw a line parallel to the given line such
that it touches the circle at one point only. It is a
tangent of the circle.
Fig. Exc_10.1_4
Exercise 10.2 (13)
In Q.1 to 3, choose the correct option and give
justification.
Question: 1 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
Solution
(a)
Let the centre of the circle be O.
Given: OQ 25 cm and PQ 24 cm
In a circle, radius is perpendicular to the tangent.
Fig. Exc_10.2_1
Thus, OP PQ
By Pythagoras theorem in OPQ,
2 2 2
OP PQ OQ
2 2
2
OP 24 25
2
OP 625 576 2
OP 49
OP 7
Hence, the radius of the circle is 7 cm.
Question: 2
If TP and TQ are the two tangents to a circle with
centre O so that , then is equal
POQ 110
PTQ
to
a.
60
b.
70
c.
80
d.
90
Fig. Exc_10.2_2
Solution (b)
The tangents to the circle are TP and TQ.
Radius of a circle is perpendicular to the tangents.
Therefore, OP TP and OQ TQ
OPT 90
OQT 90
The sum of interior angles in quadrilateral POQT is
360
OPT POQ OQT PTQ 360
90 110 90 PTQ 360
PTQ 70
Hence,
PTQ 70
Question: 3
If tangents PA and PB from a point P to a circle with
centre O are inclined to each other at angle of ,
80
then is equal to
POA
a.
50
b.
60
c.
70
d.
80
Solution
(a)
Given: and are tangents.
PA
PB Fig. Exc_10.2_3
Thus, the radius and are perpendiculars to
OB
OA
the tangents.
Thus, and .
OA PA
OB PB
and
OBP 90
OAP 90
AOBP
Sum of all interior angles
360
So,
OAP APB PBO BOA 360
90 80 90 BOA 360
BOA 360 260
BOA 100
In ,
OPBand OPA
[Tangents from a point]
AP BP
OA OB
[Common side]
OP OP
Thus, [By SSS congruence criterion]
OPB OPA
Thus, [By CPCT]
POB POA
Also,
1
POA AOB
2
100
= 50
2
Hence,
POA 50
Question: 4
Prove that the tangents drawn at the ends of a
diameter of a circle are parallel.
Solution
Fig. Exc_10.2_4
Let the diameter of the circle be .
XY
At point , two tangents are
Xand Y
PQ andRS
drawn. So, and .
OX RS
OY PQ
OXR 90 OXS
OYP 90 OYQ
Observe that,
[Alternate interior angles]
OXR OYQ
[Alternate interior angles]
OXS OYP
Alternate interior angles are equal. So, lines
are parallel.
PQ andRS
Question: 5
Prove that the perpendicular at the point of contact
of the tangent to a circle passes through the centre.
Solution
Let be the centre of circle and be a tangent
O
XY
that touches the circle at point .
P
We have to prove that the line perpendicular to
XY
at passes through the point .
P
O
We will prove this by a contradiction.
Let the perpendicular to at do not pass
XY
P
through the centre .
O
Let it is passes through another point .
Q
Join and .
OP
QP Fig. Exc_10.2_5
Perpendicular to at passes through .
XY
P
Q
Thus,
QPY 90 ...... 1
According to the figure, the centre of the circle is
O
and the point of contact is .
P
So,
OPY 90 ...... 2
By equation (1) and (2),
QPY OPY ...... 3
From the figure, observe that
QPY OPY ...... 4
Thus, is not possible.
QPY OPY
It is possible only when the lines and QP
OP
coincide.
Thus, perpendicular to XY through P, passes through
the centre O. Question: 6
The length of a tangent from a point A at distance 5
cm from the centre of the circle is 4 cm. Find the
Solution
Let O as a centre of a circle.
Fig. Exc_10.2_6
From point A, AB tangent is drawn on the circle.
Given: and
OA 5cm
AB 4 cm
In ,
ABO
.
OB AB
Apply, Pythagoras theorem in ,
ABO
2 2 2
AB BO OA
2 2 2
4 BO 5
2
16 BO 25 2
BO 9
BO 3
Thus, radius of the circle is cm.
3
Question: 7
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.
Solution
Let the point O be the centre of two concentric
circles.
Fig. Exc_10.2_7
Also, chord of the larger circle is AB and touches the
smaller circle at point P.
Thus, AB becomes the tangent to the smaller circle.
[ is radius of smaller circle]
OP AB
OP Apply Pythagoras theorem in ,
OAP
2 2 2
OP AP OA
2
2 2
3 AP 5
2
AP 25 9
2
AP 16
AP 4
In ,
OAB
OP AB
AP PB
Perpendicular from thecentre
of thecirclebisectsthechord
Now,
AB 2AP
2 4
8
Thus, length of the chord is cm.
8
Question: 8
A quadrilateral ABCD is drawn to circumscribe a
circle. Prove that . Fig. Exc_10.2_8
Solution
Observe that,
DR DS
Tangentsfrom point
D on thecircle
...... 1
CR CQ
Tangentsfrom point
C on thecircle
...... 2
BP BQ
Tangents from point
Bon thecircle
...... 3
AP AS
Tangents from point
A on thecircle
...... 4 DR CR BP AP DS CQ BQ AS
DR CR BP AP DS AS CQ BQ
Hence proved.
Question: 9
XY and are two parallel tangents to a circle with
XY
centre O and another tangent AB with point of
contact C intersecting XY at A and at B. Prove
XY
that .
AOB 90
Fig. Exc_10.2_9(Ques.)
Solution
Join point O to C. Fig. Exc_10.2_9(Sol.)
In and ,
OPA
OCA
OP OC
AP AC
Tangents from point
A on thecircle
[Common side]
AO AO
[By SSS congruence criterion]
OPA OCA
POA COA ...... 1
Similarly,
OQB OCB
QOB COB ...... 2
The diameter POQ of the circle is a straight line.
Thus,
POA COA COB QOB 180
From equations (1) and (2), POA COA COB QOB 180
COA COA COB COB 180
2 COA 2 COB 180
COA COB 90
AOB 90
Hence proved.
Question: 10
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.
Solution
Let O be the centre of the circle.
Fig. Exc_10.2_10
Let P be the external point and PX and PY are the
two tangents.
XY is the line segment. Observe that .
OX PX
Thus,
OXP 90
Similarly, .
OY PY
.
OYP 90
OXPY
OXP XPY PYO YOX 360
90 XPY 90 YOX 360
XPY YOX 360 180
XPY YOX 180
Hence, 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.
Question: 11
Prove that the parallelogram circumscribing a circle
is a rhombus.
Solution
Given: Let ABCD be a parallelogram.
AB CD ...... 1 Fig. Exc_10.2_11
Observe that,
AW AZ
Tangents from point
A on thecircle
BW BX
Tangents from point
Bon thecircle
CY CX
Tangents from point
C on thecircle
DY DZ
Tangents from point
D on thecircle
DY CY BW AW DZ CX BX AZ
DY CY BW AW DZ AZ CX BX
From equations (1) and (2), 2AB 2BC
AB BC ...... 3
Compare equations (1), (2) and (3),
AB BC CD DA
Hence, ABCD is a rhombus.
Question: 12
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.
Fig. Exc_10.2_12(Ques.)
Solution Sides AB and AC touch the circle at point E and F
respectively.
Let the length of AF be .
x
In ,
ABC
CF CD 6cm
Tangents from point
C on thecircle
BE BD 8cm
Tangents from point
Bon thecircle
AF AE x
Tangents from point
A on thecircle
Fig. Exc_10.2_12(Sol.)
Now,
AB AE EB x 8
BC BD DC
8 6
14 CA CF FA
6 x
By the Hero’s formula,
2s AB BC CA
2s x 8 14 6 x
2s 28 2x
s 14 x
ABC s s a sArea o b sf c
14 x 14 x 14 14 x 6 x 14 x 8 x
14 x 14 x 14 14 x 6 x 14 x 8 x
14 x x 8 6
2
4 3 14x x
1
OBCAre Oa
2
f Do BC
1
4 14
2
28
1
OCAAre Oa
2
f Fo AC
1
4 6 x
2
12 2x 1
OABAr OE AB
2
ea of
1
4 8
2
x
16 2x
Now, find the value of .
x
Area of ABC Area of OBC
Area of OCA
Area of OAB
2
4 3 14 28 12 2 16 2x x x x
2
4 3 14 56 4x x x
2
3 14 14x x x
2
2
3 14 14x x x
2 2
42 3 196 28x x x x
2
2 14 196 0x x
2
7 98 0x x
2
7 98 0x x
2
14 7 98 0x x x
14 7 14 0x x x
14 7 0x x
14,7x
Length is always positive so, is rejected.
14 Thus,
7x
Hence, cm
AB 8 7 8 15x
cm.
CA 6 6 7 13x
Question: 13
Prove that the opposite sides of a quadrilateral
circumscribing a circle subtend supplementary
angles at the centre of the circle.
Solution
Let O be the centre of circle and quadrilateral ABCD
touches the circle at points .
W X Y Z, , ,
Fig. Exc_10.2_13
Join the vertices of the quadrilateral with the centre
of the circle.
In and ,
OAW
OAZ
AW AZ
Tangents from point
A on thecircle
[common side]
OA OA
[SSS congruence criterion]
OAW OAZ
Thus,
WOA AOZ
1 8
Similarly,
2 3
4 5
6 7
Now,
1 2 3 4 5 6 7 8 360
1 8 2 3 4 5 6 7 360    
2 1 2 2 2 5 2 6 360
2 1 2 2 5 6 360
1 2 5 6 180
AOB COD 180
AOB COD 180
Similarly,
BOC DOA 180
Thus, the opposite sides of a quadrilateral
circumscribing a circle subtend supplementary
angles at the centre of the circle.