# Free Circles 01 Practice Test - 9th Grade

A tangent is drawn at a point P on a circle. A line through the centre O of a circle of radius 7 cm cuts the tangent at Q such that PQ = 24 cm. Find OQ.

A.

10 cm

B.

15 cm

C.

20 cm

D.

25 cm

#### SOLUTION

Solution : D

Since tangent at a point on the circle is perpendicular to the radius through that point of contact. OPOQ

In right angled triangle OPQ,

OQ2=OP2+PQ2 (By Pythagoras theorem)

OQ2=72+242

OQ2=49+576

OQ2=625

OQ=625=25 cm

ABCD is a cyclic quadrilateral whose side AB is a diameter of the circle through A, B, C and D. If ADC=130, find BAC.

A.

(1202)

B.

(1203)

C.

60

D.

40

#### SOLUTION

Solution : B and D

Given: ABCD is a cyclic quadrilateral whose side AB is the diameter of the circle and ADC=130.

To find BAC, D+B=180(Opposite angles of a cyclic quadilateral )

130+B=180

B=180130=50

ACB=90(Angle in a semicircle)

In ΔABC,

BAC+50+90=180(Since sum of angles of a triangle is 180)

BAC=1809050=40

An equilateral triangle ABC is inscribed in a circle with centre O. Then, BOC is equal to ___. A. 30
B. 60
C. 90
D. 120

#### SOLUTION

Solution : D

Given that ΔABC is equilateral.
BAC=60
Since the angle subtended by a chord at the centre of a circle is twice the angle subtended by the same chord at any other point on the remaining part of the circle, we have
BOC=2 BAC=2×60=120.

ABCD is a cyclic quadrilateral. If BCD=100  and ABD=70, then find ADB. A. 30
B. 40
C. 50
D. 60

#### SOLUTION

Solution : A

Given that ABCD is a cyclic quadilateral such thatBCD=100 and ABD=70. Since opposite angles of a cyclic quadrilateral are supplementary,
DAB+DCB=180.

DAB=180DCB
=180100=80

Consider ADB, since sum of angles of a triangle is 180, we have

There are only ___ ways in which a circle and line can interact.

#### SOLUTION

Solution : Circle and line can interact in maximum three ways:
1. When a line is neither touching or cutting a circle.
2. When a line is touching a circle.
3. When a line is cutting a circle.

A line can be taken as a circle with infinite radius and a circle can be assumed to be a polygon with infinite sides.

A. True
B. False

#### SOLUTION

Solution : A

Draw an arc of smaller radius and join its ends at its centre. Now draw an arc of bigger radius and try to join its ends at its centre, now draw a straight line and try to join its ends at its centre we can see that it will meet its centre at infinity, so from this we can conclude that a line is a circle of infinite radius. Take a triangle and add it to another triangle it will form a quadrilateral, add another triangle to this quadrilateral it will form a pentagon, keep adding triangles to this polygon at one point you will realize that a polygon with infinite sides will be a circle, with each side being infinitesimally small.

In the given circle, with AB as a diameter, find the value of x. A.

50

B.

40

C.

30

D.

60

#### SOLUTION

Solution : D ABD=ACD=30[Angles in the same segment]
x+90+30=180
x=(180120)=60
Hence,x=60

If  ABCD is a cyclic quadrilateral with AB as the diameter of the circle and ACD=50, then BAD = ___.

A.

70

B.

80

C.

50

D.

40

#### SOLUTION

Solution : D Since an angle in a semicircle is a right angle, ACB=90.
But, BCD=ACD+ACB=50+90=140.
Consider the cyclic quadrilateral ABCD. Since opposite angles are supplementary,
BCD+DAB=180.

A circle with center O and radius 5 cm has two chords AB and AC, such that AB = AC = 6 cm. If by joining B and C the line segment passes from the center of the circle O, then what is the length of BC?

A.

5 cm

B.

9 cm

C.

10 cm

D.

11.9 cm

#### SOLUTION

Solution : C

As the line segment BC passes through the center, it means that it is a diameter of the circle.

Given that radius of the circle = 5 cm
Diameter = 5 + 5 = 10 cm

AB and CD are two parallel chords of a circle such that AB = 10 cm and CD  = 24cm.  If the chords are on the opposite sides of the centre and the distance between them is 17 cm, then the radius of the circle is _____.

A. 14 cm
B. 10 cm
C. 13 cm
D. 15 cm

#### SOLUTION

Solution : C Given AB and CD are two chords of a circles on opposite sides of the centre.

Construction: Draw perpendiculars OE and OF onto AB and CD respectively from centre O.
AE = EB = 5cm and CF = FD = 12 cm
[ Perpendicular drawn to a chord from center bisects the chord]

Given,
Distance between two chords = 17 cm
Let distance between  O and F =x cm
And distance between  O and E  =(17x) cm

In  ΔOEB,
OB2=OE2+EB2
[Pythagoras theorem]
=(17x)2+52 ---(1)

In  ΔOFD,
OD2=OF2+FD2
[Pythagoras theorem]
=(x)2+122----------→(2)

But OB = OD ( radii of the same circle).

From 1 & 2,
(17x)2+52=(x)2+122
289+x234x+25=x2+144
34x=170
x=5
Subsitute x in equation (2);
OD2=(5)2+122=169
OD=13
∴ Radius of the circle is 13 cm.