# Free Circles 02 Practice Test - 10th Grade

### Question 1

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

#### 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=625−576=49

⇒OP=7 cm

### Question 2

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∘.

True

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=120∘–90∘

∠OPQ=30∘

### Question 3

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.

2 cm, 3 cm, 4 cm

3 cm, 4 cm, 1 cm

1 cm, 2.5 cm, 3.5 cm

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=9⟹z=4

Now using, (3)⟹x=7−z=7−4=3

And (2.1)⟹y=6−z=6−4=2Therefore, the radii of the circles are 3 cm, 2 cm and 4 cm.

### Question 4

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

### Question 5

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.

√a2−b2

√a2+b2

2√a2−b2

2√a2+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=√OA2−OC2=√a2−b2

∴ AB=2AC=2√a2−b2 [ ∵ C is the midpoint of AB]i.e., Length of the chord AB=2√a2−b2

### Question 6

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

#### 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

### Question 7

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?

60∘

50∘

20∘

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∘

### Question 8

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

AD=AB+BC+CA

2AD=AB+BC+CA

3AD=AB+BC+CA

4AD=AB+BC+CA

#### SOLUTION

Solution :B

We know thatAD=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

### Question 9

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.

True

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.

∴ PA⊥OA, i.e., ∠OAP=90∘ . . . (i)

and PB⊥OB, 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∘].

### Question 10

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?

This situation holds no signifiance

OP needs be the radius of circle

OP is perpendicular to XY

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.