# QUESTIONS ON CIRCLE (PART-1)

## QUESTIONS ON CIRCLE (PART-1)

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 1

**AC and BC are two equal chords of a circle. BA is produced to any point P and CP when joined cuts circle at T. Then**

A) CT : TP = AB : CA

B) CT : TP = CA : AB

C) CT : CB = CA : CP

D) CT : CB = CP : CA

** RONNIE BANSAL**

Draw the figure according to the question. Now concentrate on the two triangles CPA & CAT. We have to choose these two triangles only as the ratio of the sides of only these triangles is given in the answer options. CB too is there but it’s equal to CA only.

Well, now see if the angles of these triangles could be proved equal. As the quadrilateral CTAB is cyclic, angles CTA+CBA=180; and angles CTA+ATP also equal to 180 (angles on straight line), so ∠ATP = ∠CBA.

Now you can easily show that ∠CAP= ∠CTA. ∠PCA= ∠ACT (same angle). So ∠CAP= ∠CAT (remaining angles}. From above you can see that these two triangles are similar. Now think over a bit, you’ll find option ‘C’ as correct.

QUERY 2

** P & Q are mid-points of two chords AB and AC of a circle. Center is O, OP and OQ are extended to meet the circle at R & S respectively. T is a point anywhere on the major arc RS of the circle. ∠BAC = 32°. Find ∠RTS.**

A) 32

B) 74

C) 106

D) 64

** RONNIE BANSAL
** Concentrate on the quadrilateral OPAQ. In this ∠OPA = ∠OQA = 90° (both P & Q are mid points of two chords and they join the centre O.

So their sum equals 180°. Means the sum of the two remaining angles of the quadrilateral OPAQ i.e. ∠PAQ and ∠POQ is also 180°. But ∠PAQ=32° (∠PAQ and ∠BAC are the same). So ∠POQ=180 - 32 = 148.

But angles POQ and ROS are same; therefore ∠ROS = 148. Therefore ∠RTS = 148/2 = 74 (angle subtended by the arc RS on the remaining part of the circle) [option ‘B’]

THEOREM: 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.

QUERY 3

**In clock between 5 and 6 when hour and minute hands will stick together?**

A) 5 hour 25 min

B) 5 hours 300/11 min

C) 5 hr 289/11 min

D) 5 hr

** Vijay Bharath Reddy**

We get the formula, (11/2) x m – 30 x h = 0

=> (11/2) x m – 30 x 5 = 0

=> m = 300/11

Therefore, the time required is 5 hr 300/11 min (option ‘B’)

**Sumanashis**

AT 5.00 Angle = 150°. Now with the every 1′, the difference is reduced by (11/2)°. Hence to stick together difference must be 0°.

NOW, (11/2)° in 1′, Hence 150° in 150*2/11 = 300/11 min. So required time = 5 hrs + 300/11 min (option ‘B’)

QUERY 4

**O is the center of a circle and OX and OY are two radii; ∠XOY = 90 and area of ΔXOY is 32 cm². Find the Area of the circle.**

A) 60

B) 64π

C) 32π

D) 64

**RONNIE BANSAL
**The ΔXOC is a right angle triangle which is isosceles also where OX = OC (each a radius of the same circle). We know area of an isosceles right triangle is side

^{2}× 1/2; where side is the

But area is 32 cm

^{2}

Therefore radius^{2} × ½ = 32

=> radius=8

Now the area of the circle=π8^{2
}=64π (option ‘B’)

QUERY 5

**A, B, C are three points on a circle. The tangent at A meets BC produced at T, ∠BTA=40, ∠CAT=44. The angle subtended by BC at the centre of the circle is ?**

A) 84

B) 92

C) 96

D) 104

**Amit jha**

Given ∠CAT = 44° and ∠CTA = 40°

We have ∠CAT = ∠ABC = 44° (alternate segment of the circle)

Again ∠BCA = ∠CAT + ∠CTA = 44 + 40 = 84° (exterior angle of a triangle)

Now ∠ABC + ∠BCA + ∠CAB = 180°

=> 44° + 84° + ∠CAB = 180°

=> ∠CAB = 52°

Again ∠BOC = 2∠CAB = 2 × 52 = 104° (angle subtended by a chord on the center is double to the angle subtended on the circle) [option ‘D’]

QUERY 6

**A, B, C are 3 points on the circumference of a circle. Chord AB forms 90° at the centre and chord AC forms 110°. Find the ∠BAC.
**

A) 160°

B) 180°

C) 80°

D) 100°

**MAHA GUPTA
** Let the center of the circle be ‘O’; then ∠AOB + ∠AOC + ∠BOC = 360

=> 90 + 110 + ∠BOC = 360

=> BOC = 160

But ∠BAC is subtended on the other part of the arc BOC.

So ∠BOC = 2∠BAC (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.

=> ∠BAC = 160/2 = 80° (option ‘C)

QUERY 7

**The area of circle whose radius is 8 cm is trisected by two concentric circles. The ratio of radii of the concentric circles in ascending order is ?**

A) 1, 2, 3

B) 2, 3, 5

C) 1, √2, √3

D) √2, √3, √5

**Vijay Bharath Reddy
**Let the radii be a, b, c in ascending order

Since the area is trisected, therefore πa^2 = π(b² – a²) = π(c² – b²)

From 1 & 2

b =√2a

From 2 & 3

c = √3a

=> a :b : c = a : √2a : √3a

= 1 : √2 : √3 (dividing all by ‘a’) Option ‘C’

QUERY 8

**If the center of a circle passing through points A, B, C makes ∠BAO = 30° and ∠BCO = 40°; find ∠AOC.**

A) 240

B) 210

C) 220

D) 120

**RONNIE BANSAL
**As the circle passes through A, B, C, so according to the question BA and BC are two chords of the circle with B as the common point.

∠BAO = 30 & ∠BOC = 40 (given)

Therefore ∠ABC = ∠BAO + ∠BOC = 30+40 = 70 (Angle at the meeting point of two chords of the same circle is the sum of angles made by their end points with the center)

Again ABCO is a quadrilateral with ∠ABC & ∠AOC, and ∠BAO & ∠BCO as two opposite pairs.

Therefore ∠ABC + ∠AOC = 360 – 70 = 290

=> ∠AOC = 290 – ∠ABC

= 290 – 70 = 220 (option ‘C).

NOTE: If 220 is not in the answer options the desired angle will be 360 – 220 = 140

QUERY 9

**Two circle of 15 cm and 20 cm intersect. Distance between the two centers is 25 cm. The length of common chord is ?**

A) 20

B) 24

C) 15

D) 12

**Vijay Bharath Reddy
**Let the centres of circles be A and B; D being the intersecting point of the radii and the two circles meet the circles at P and Q; therefore PQ is the common chord.

In this way we’ll have a triangle with sides 20, 15 and 25

Now the area of the triangle =√{s(s-a)(s-b)(s-c)}; where ‘s’ is half of the sum of the sides of the triangle

=> √[30(30 – 20)( 30 – 15)(30 – 25)]

= 150

But the common chord intersects the line joining the centres of the circles; therefore PD is the height of this ∠APB

Thus the area of the ∠APB is also equal to 1/2 (25*h) [Area of a triangle also = ½base *height]

Therefore 25/2h = 150

h = 12

But this height is half of the chord

Hence length of the chord = 2*12 = 24 (option ‘B’)

QUERY 10

**The lengths of two chords of a circle intersecting each other perpendicularly are 6 cm and 8 cm respectively. If the point of intersection is at a distance of 5 cm from the center; find the radius of the circle ?**

A) 7 cm

B) 8 cm

C) 10 cm

D) 5 cm

**Vijay Bharath Reddy
**Draw perpendicular ‘m’ from the center of the circle on 8 cm chord and ‘n’ on 6 cm chord and let ‘r’ be the radius of the center. Join center of the circle to one of the end point of 8 cm chord and to one of the end point of 6 cm chord.

In this way we’ll have two right triangles; one with sides ‘m’ (1/2 of 8) = 4 cm and radius as diagonal and the other with sides ‘n’ (1/2 of 6) = 3cm and radius as diagonal.

Now using pythagoras

r² = m² + 4² ……..i

r² = n² + 3² …….ii

Adding (i) and (ii)

2r² = 4² + 3² + m² + n² …….iii

=> 2r² = 5² + (m² + n²)

If we see carefully we’ll also have a rectangle whose length and breadth will be ‘n’ and ‘m’ respectively and the line joining the intersecting point of the chords to the center as one of its diagonals.

Therefore m² + n² = 5²

Now putting this in equation (iii)

2r² = 5² + 5²

=> 2r² = 2*5²^2

=> r = 5 (option ‘D’)