# Free Circles 02 Practice Test - 10th Grade

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

#### SOLUTION

Solution : Given that PQ = 24 cm and OQ = 25 cm.

To find OP, apply Pythagoras theorem for OPQ.

OP2+QP2=OQ2

OP2+242=252

OP2=625576=49

OP=7 cm

State true or false.
PQ is a tangent drawn from a point P to a circle with center O and QOR is a diameter of the circle such that ∠POR = 120 , then ∠OPQ is 30.

A.

True

B.

False

#### SOLUTION

Solution : A Given that POR=120

We know that OQP is 90

Through external angle theorem POR=OQP+OPQ

120=90+OPQ

OPQ=12090

OPQ=30

Three circles touch each other externally. The distance between their centres is 5 cm, 6 cm and 7 cm. Find the radii of the circles.

A.

2 cm, 3 cm, 4 cm

B.

3 cm, 4 cm, 1 cm

C.

1 cm, 2.5 cm, 3.5 cm

D.

1 cm, 2 cm, 4 cm

#### SOLUTION

Solution : A

Consider the below figure wherein three circles touch each other externally. Since the distances between the centres of these circles are 5 cm, 6 cm and 7 cm respectively, we have the following set of equations with respect to the above diagram:
x+y = 5     …..(1)

y+z = 6     ......(2) ( y=6-z)...   (2.1)

x+z = 7     …..(3)

Adding (1), (2) and (3), we have 2(x+y+z)=5+6+7=18

x+y+z=9....(4)

Using (1) in (4), we have 5+z=9z=4

Now using, (3)x=7z=74=3

And (2.1)y=6z=64=2

Therefore, the radii of the circles are 3 cm, 2 cm and 4 cm.

In the given figure, a circle touches the side BC of ΔABC at P and AB and AC produced at Q and R respectively. If AQ = 15 cm, find the perimeter of ΔABC. __

#### SOLUTION

Solution :

We know that

AQ = AR

AQ = AC + CR

But CR = CP

Therefore AQ = AC + CP    …..(i)

Also AQ = AB + BQ

But BQ = BP

AQ = AB + BP     .….(ii)

Adding equations (i) and (ii)

2AQ = AC + CP + BP + AB

2AQ = Perimeter of ΔABC

Perimeter of ΔABC = 2 × 15 = 30 cm

Two concentric circles of radii a and b (a > b) are given. Find the length of the chord of the larger circle which touches the smaller circle.

A.

a2b2

B.

a2+b2

C.

2a2b2

D.

2a2+b2

#### SOLUTION

Solution : C

Let O be the common centre of the two circles and AB be the chord of the larger circle which touches the smaller circle at C.
Join OA and OC.
Then OC AB
Let OA = a and OC = b. Since OC AB, OC bisects AB
[ perpendicular from the centre to a chord bisects the chord].

In right Δ ACO, we have

OA2=OC2+AC2    [by Pythagoras' theorem]

AC=OA2OC2=a2b2
AB=2AC=2a2b2     [ C is the midpoint of AB]

i.e., Length of the chord AB=2a2b2

In the given figure, the circle touches the sides AB, BC, CD and DA of a quadrilateral ABCD at the points P, Q, R, S respectively. If AB = 11 cm, BC =x cm, CR = 4 cm and AS = 6 cm, the value of x is

___ cm. #### SOLUTION

Solution :

AP = AS = 6 cm

BP = BA – AP = 11 – 6 = 5

BQ = BP = 5

CQ = CR = 4

x = CQ + BQ

x = 4 + 5 = 9 cm

ABC is an isosceles triangle and AC, BC are the tangents at M and N respectively. DE is the diameter of the circle. ADP = BEQ = 100. What is value of PRD? A.

60

B.

50

C.

20

D.

Can't be determined

#### SOLUTION

Solution : C

ADB is a straight line. By linear pair axiom,
ADP + PDB = 180
100 + PDB = 180
PDB = 80
Similarly QED = 80

We have, ∠DPE = 90 (angle subtended by a diameter)

In DPE,
DPE + PED + EDP = 180
[Angle sum property of a triangle]
∠PED = 10
Similarly QDE = 10

In DRE,
DRE + RDE + RED = 180
[Angle sum property of a triangle]
∠DRE = 160

PRE is a straight line. By linear pair axiom,
PRD + DRE = 180
PRD + 160 = 180
PRD = 20

In the adjoining figure, if AD, AE and BC are tangents to the circle at D, E and F respectively. Then A.

AD=AB+BC+CA

B.

2AD=AB+BC+CA

C.

3AD=AB+BC+CA

D.

4AD=AB+BC+CA

#### SOLUTION

Solution : B We know that

AD=AE

AD=AB+BE

Since BE=BF as tangents drawn from an external point to a circle are equal , AD=AB+BF……(1)

Also

AD=AC+CD

AD=AC+CF  ……(2)    (CD and CF are tangents drawn from the external point C to the circle)

Adding equation (1) and (2),

AD+AD=AB+BF+CF+AC

2AD=AB+BC+AC

State True or False.

PA and PB are tangents to the circle with centre O from an external point P, touching the circle at A and B respectively. Then the quadrilateral AOBP is cyclic.

A.

True

B.

False

#### SOLUTION

Solution : A Given PA and PB are tangents to the circle with centre O from an external point P.

We know that the tangent at any point of a circle is perpendicular to radius at the point of contact.

PAOA, i.e.,  OAP=90      . .  . (i)
and  PBOB, i.e.,  OBP=90         . .. (ii)

Now, the sum of all the angles of a quadrilateral is 360

AOB+OAP+APB+OBP=360
AOB+APB=180  [using (i) and (ii)]

quadrilateral OAPB is cyclic   [since both pairs of opposite angles have the sum 180].

O is the centre of the circle and line XY shares one common point with the circle. A line segment OP is drawn from O to line XY such that P is a point on line XY. What happens if P lies on the circle?

A.

This situation holds no signifiance

B.

OP needs be the radius of circle

C.

OP is perpendicular to XY

D.

The angle between OP and XY is variable

#### SOLUTION

Solution : B and C

XY touches the circle at only one point. Thus, it is a tangent to the circle. The tangent touches the circle at only one point which lies on the circumference (point P).

Thus, any other point on line XY other than P lies outside the circle. Let such a point be Q. Now let us compare lengths OP and OQ. It is clear that OP is equal to radius since P lies on circle. Q lies outside. Thus, OP is lesser than OQ. A point which is outside the circle will obviously be farther away from circle centre than a point which lies on the circle.

It can be seen that at any position of point Q on line XY except on P, OP is lesser than OQ. In other words, OP is the smallest distance between O and XY. The smallest distance between a line and a point is the perpendicular distance. Hence OP is perpendicular to XY. 