# QUESTIONS ON CIRCLE (PART-3)

#### QUESTIONS ON CIRCLE (PART-3)

Most of these questions are taken from the previous year’s examinations conducted by the Staff Selection Commission (SSC) on Geometry (Circle) of the Quantitative Aptitude (Maths) section of the following exams as well as other similar exams. They are all solved and supported by detailed explanation.

1. Combined Graduate Level (CGL) Exam Tier-I & Tier-II

2. Combined Higher Secondary (10+2) Exam (CHSL) Tier-I

3. SI in Delhi Police and CPO Exam Paper-I & Paper-II

#### QUERY 21

**Two circles of radii 8 and 2 cm touch each other externally at point A. PQ is direct common tangent. Length of QP is?**

A) 8 cm

B) 7.5 cm

C) 7 cm

D) 9 cm

Draw QR || ST

So, QR = ST = 10 cm

Now in right triangle QPR

PR = PS – RS = 8 – 2 = 6 cm

Diagonal QR = 10 cm

Therefore by Pythagoras, QP = √(10² – 6²) = 8 cm (option ‘A’)

QUERY 22

**One chord of a circle is known to be 10.1 cm. The radius of this circle must be?**

A) 5 cm

B) greater than 5 cm

C) greater than or equal to 5 cm

D) less than 5 cm

**MAHA GUPTA**

The longest chord of the circle is its diameter

Means the diameter cannot be bigger than 10.1 cm, Now letting the diameter of the circle be 10.1 cm

Therefore the radius = 10.1/2 = 5.5 cm

Hence option ‘B’ is correct

QUERY 23

**P is a point outside a circle and is 13 cm away from its center. A secant drawn from the point P intersects the circle at points A and B in such a way that PA = 9 cm and AB = 7 cm. The radius of the circle is?**

A) 5 cm

B) 4 cm

C) 4.5 cm

D) 5.5 cm

(Option ‘A’)

#### QUERY 24

**In a circle, AB is diameter, CD is chord of length equal to radius of the circle. AC and BD when produced intersect at point E. Find ∠AEB?**

A) 60°

B) 65°

C) 70°

D) 55°

**MAHA GUPTA
**

Triangle ODC is equilateral. (OC and OD being the radii of the circle and CD being given equal to the radius)

Therefore, ∠COD = 60°

Now, ∠CBD =½∠ COD (The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle. Here arc CD subtending ∠CBD at the center and ∠COD on any other part of the circle)

=> ∠CBD = 30°

Now ∠ACB = 90° (angle in the semi-circle)

So, ∠BCE = 180° – ∠ACB = 90°

Therefore ∠CEB = 90° – 30° = 60°, i.e. ∠AEB = 60° (option ‘A’)

QUERY 25

A) 80°

B) 85°

C) 70°

D) 75°

**MAHA GUPTA**

The arch RS is subtending the ∠ROS = 40 at the centre and the ∠RQS an other angle on the remaining part of the circle; therefore ∠RQS = ½∠ ROS (The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.)

=> ∠RQS = 40/2 = 20°

Now ∠PRQ = 90° (angle in the semi-circle)

So, ∠QRT = 180° – ∠PRQ = 90°

Therefore ∠RTQ = 90° – 20° = 70°, i.e. ∠RTS = 70° (option ‘C’)

QUERY 26

**O is centre of a circle. Two cords AC and BD intersect at P. If ∠AOB =15 and ∠APB=30, find tan² ∠APB + Cot² ∠COD**

A) 4/3

B) 1/3

C) 2/3

D) 1

As ∠AOB = 15, hence ∠ADB = 15/2

∠APD = 150

Now in ΔAPD, ∠P = 150, ∠D = 15/2

Therefore ∠A = 45/2 and ∠DOC = 45

Then tan² 30 + cot² 45 = 1/3 + 1 = 4/3 (option ‘A’)

#### QUERY 27

**Two circles touch each other at a point A and PQ is a common tangent to both the circles; find ∠PAQ.**

A) 60°

B) 75°

C) 80°

D) 90°

**SHIV KISHOR**

If two circles, either equal or not, when touch at a point, say A, their common tangent, say PQ always makes angle PAQ = 90 (option ‘D’)

QUERY 28

**Radii of two circles are 7 cm and 3 cm respectively. If one of those circles is completely inside the other circle, what will be the distance between the centers?**

A) 4 cm more

B) more than 5 cm

C) less than 4 cm

D) none of the above

**MAHA GUPTA**

Distance between the centers of two circles is the difference of their radii if one of them is touching the other internally.

But here the circles are not touching, rather one of them lies completely inside of the other, therefore the distance between the centers will be less than the difference of their radii, means less than (7 – 3) i.e. 4 (option ‘C’)

QUERY 29

**ABC is an isosceles triangle with AB = AC. A circle through B touching AC at the middle point intersects AB at P. Then AP : AB is?**

A) 4 : 1

B) 2 : 3

C) 3 : 5

D) 1 : 4

**SHIV KISHOR
**

AM is a tangent to the circle and APB is a secant to it. Thus by tangent theorem AM² = AP × AB

But AM = AC/2 = AB/2

=> AB²/4 = AP × AB

=> AP/AB = 1/4 = 1 : 4 (option ‘D’)

#### QUERY 30

**AB = 2r is the diameter of a circle. If a chord CD intersect AB at right angle at point P in the ratio 1 : 2, then CD is ?**

A) 4√2r/3

B) 2√2r/3

C) 3√2r/4

D) √2r/3

**SHIV KISHOR
**

AB = 2r

The chord CD intersect AB (2r) at point P in the ratio 1 : 2

Means AP : PB = 1 : 2

=> AP = 2r/3

OP = OA AP

=> OP = r – 2r/3

=> OP = r/3

(Option ‘A’)