MCQ: Circles - 1 - SSC CGL MCQ

In ΔACP
CP = CD – PD = 25 – 18 = 7

Similarly, PB = 24 cm
∴ AB = AP + PB
= 24 + 24
= 48 cm.
Hence, option D is correct.

According to question , we draw a figure of two equal circles of radius 4 cm intersect each other such that each passes through the centre of the other ,

OC = 2cm
OA = 4cm
∴ AC = √OA² - OC²
∴ AC = √4² - 2² = √16 - 4
AC = √12 = 2√3
∴ AB = 4√3cm

Given, one chord AB = 24 cm
Then, AE = EB = 12 cm
Diameter = 30 cm ⇒ radius, AO = OC = 15 cm
From ΔAOE, By pythagoras theorem

Distance between two chords, EF = 21 cm (given)
∴ OF = EF – OE = 21 – 9 = 12 cm
From ΔCOF, By pythagoras theorem

Hence, option B is correct.

So,
length of perpendicular drawn from center
= 15 – 10 = 5 cm.
Hence, option D is correct.

The largest chord of a circle is its diameter. So,

Hence, option B is correct.

Given, AB = 16 cm and OO' = 12 cm
∴ AC = CB = 8 cm and OC = CO' = 6 cm
From ΔAOC, By pythagoras theorem

Hence, option A is correct.

In ΔABC,
AB = BC = AC
Hence, ΔABC is equilateral triangle.
Let r be the radius of incircle and R be the radius of circumcircle.

= 2 : 1.
Hence, option A is correct.

Chord, AB = 8 cm
Then, AC = CB = 4 cm
Perpendicular distance between centre and chord,
OC = 3 cm

Hence, option B is correct.

Given, PD = 5 cm Then, PC = PD – CD = 5 – 3 = 2 cm
Similarly, PA = (PB – 6) cm
Note : If two chords AB and CD of a circle intersect inside or outside the circle when produced at a point P, then
PA x PB = PC x PD ⇒ (PB – 6) x PB = 2 x 5 ⇒ PB2 – 6PB – 10 = 0
By Sridharacharya formula,

Hence, option B is correct.

By definition of tangent,
A tangent to a circle is straight line that touches the circle at a single point. Also, tangent at the
end points of a diameter of a circle is perpendicular to the diameter.
So, both statements are correct.

Hence, option C is correct.

AO = OB = AB
⇒ ∠AOB = 60° [∵ ΔAOB is equilateral]
Note : The angle subtended by an arc of a circle at the centre is double the angle subtended by it
at any point on the remaining part of the circle.
∴ ∠ACB = 30°
Hence, option A is correct.

Given, ∠ACB = 65°
Note : The angle subtended by an arc of a circle at the centre is double the angle
subtended by it at any point on the remaining part of the circle.
∴ ∠AOB = 2 × 65° = 130°
Note : A tangent at any point of a circle is perpendicular to the radius through
the point of contact.
∴ ∠OAP = 90°

We know that, the sum of the three angles of a triangle is 180°.
∴ ∠APO = 180° – 90° – 65° = 25°
Hence, option A is correct.

OB = OA = radius

and ∠OAB = ∠OBA = 60° So, ΔAOB is an equilateral triangle. Then, AB = 5 cm
So, Area, x = Area of circle – Area of hexagon

Hence, option A is correct.

Given, Chords AB = 8 cm and CD = 6 cm
Then, AE = EB = 4 cm and CF = FD = 3 cm
EF = 1 cm
Let OE = x cm
Then, OF = (x + 1) cm
OA = OC = r cm (∵ radius)
From ΔOAE, By pythagoras theorem

From ΔOCF, By pythagoras theorem

By equation (ii) – (i),

∴ From equation (i),
9 = r² – 16
⇒ r² = 25
⇒ r = 5 cm
Hence, option A is correct.

Given, Chords AB = 10 and CD =24 cm
∴ AE = EB = 5 cm and CF = FD = 12 cm
Radius AO = OC = 13 cm
From ΔAOE, By pytharoas theorem

From ΔCOF, By pytharoas theorem

Hence, option A is corret.