MCQ: Circles - 2 - SSC CGL MCQ

∠APB = 90°
AB = Diameter = Hypotenous of triangle APB
As, the angle of semicircle is right angle
so, the circumcentre lies on midpoint of hypoteneous
Hence, option (C) is correct.

∠AOB = 48°

(As angles made by same arc AB)
Given AC and OB intersect each other at right angle.
∠ CQB = 90°
∠ CBQ = 180° – (90° + 24°) = 66°
so , ∠ OBC = 66°
Hence, option B is correct

Given, Common chord AB = 24 cm
Then, AD = DB = 12 cm
Diameter of circle of centre O = 30 cm, Then radius OA = 15 cm
And, Diamter of circle of centre O' = 26 cm, Then radius O'A = 13 cm
From ΔOAD, By pythagoras theorem

From ΔO'AD, By pythagoras theorem

Hence, option B is correct.

A unique circle can always be drawn through x number of given non-collinear points, then x must
be.
Hence, option B is correct.

BD = 50 cm
Let, BC = x cm, then CD = (50 – x) cm
In ΔABC, By pythagoras theorem

In ΔACD, By pythagoras theorem

Hence, option D is correct.

Given, Chord AB = Chord CD = 8 cm
Then, AP = PB = 4 cm and Diameter = 10 cm, then radius = 5 cm
Note : Equal chords of a circle (or of congruent circles) are equidistant from the centre.
∴ OP = OQ
From ΔOAP, By pythagoras theorem

Hence, option A is correct.

AB = 10 cm and CD = 24 cm
∴ AE = EB = 5 cm and CF = FD = 12 cm
EF = 17 cm
Let, EO = x cm, then OF = (17 – x) cm
In ΔAOE, By pythagoras theorem

In ΔCOF, By pythagoras theorem

52 + x2 = 122 + (17 – x)2
25 + x2 = 144 + 289 – 34x + x2
34x = 408
x = 12

Hence, option C is correct.

AB = 8 cm ⇒ AC = 4 cm
OA = 5 cm
In ΔAOC, By pythagoras theorem

Hence, option A is correct.

∠ PQS = 60°
∠QO'R = 130°

⇒ ∠ QRP = 180° – 60° – 65° = 55°
⇒ ∠PO'Q = 110°
In Δ QO'R
QO' = O'R
⇒ ∠O'QR = ∠O'RQ = 25°
∵ ∠O'QR + ∠O'RQ = 50°
⇒ ∠PQO' + ∠QPO' = 35°
∵ ∠PQO' + ∠QPO' = 70°
Similarly, ∠O'PR = 30°
∴ ∠RPS = 35°
Hence, option (B) is correct.

AB = 30 cm and CD = 16 cm
∴ AE = EB = 15 cm and CF = FD = 8 cm
Radii, OA = OC = 17 cm
In ΔAOE, By pythagoras theorem

Again In ΔCOF, By pythagoras theorem

Distance between chords, EF = OF – OE = 15 – 8 = 7 cm
Hence, option B is correct.

∠BAC = 90°
As, BC is the diameter of the circle.

Hence, option D is correct.

O'A = O'C = O'D = 3 cm [∵ radii of a circle]
OA = OB = 2 cm
AC = 2 x OA' = 2 x 3 = 6 cm
AB = 2 x OA = 2 x 2 = 4 cm
BC = AC – AB = 6 – 4 = 2 cm
∴ O'B = O'C – BC = 3 – 2 = 1 cm
In ΔBDO', By pythagoras theorem

Hence, option D is correct.

Let radius be 'r' and ∠POS = x°
ΔOQR isosceles ∴∠QOR = 30°
∴ ∠OQR = 120° (Sum of all angles of ΔOQR = 180°)
∴ ∠OQP = 60° (Supplementary angle)
ΔOPQ isosceles since OP = OQ = r
∴ ∠OQP = 60° = ∠OQP
∴ ∠ POQ = 60° = [Sum of all angle of Δ = 180° ]
Now SOR is a straight line
∴ x + 60° + 30° = 180°
∴ x = 90°
Hence, option (C) is correct.

In Δ AOD :
OA = OD (radius)
∠ AOD = 90 ( as OD is perpendicular to AB )
So Δ AOD is isosceles having OA and OD sides equal and one angle as 90
So the remaining wo angles are 45 each
Hence ∠ BAD = 45°
Therefore, option (B) is correct.

We know that,
length of arc = Θ × radius

Hence, option B is correct.