Math Olympiad Test: Circles- 1 - Class 9 MCQ

The largest chord of a circle is called diameter of the circle.

Any chord which crosses the centre of the circle is the largest chord of the circle, which is the diameter of the circle.

Let the radius of the circle be r, and the length OP be x

∴ In ∆OPB, 
OP² + PB² = OB² = r²
⇒ x² + (4)² = r²   [∵ PB = AB/2]
⇒ x² = r² – 16 ….(i) 
In ∆OCQ,
OQ² + CQ² = r²
⇒  (7 – x)² + (3)² = r²
[∵ OQ = PQ - OP = 7 - x]
⇒ (7 – x)² = r² – 9 ….(ii)
Subtracting (ii) from (i), we get
x² – (7 – x)² = – 7
⇒ (x – 7 + x) (x + 7 – x) = – 7 
⇒    (2x – 7) (7) = – 7  
⇒  (2x – 7) = – 1 
⇒ x = 3
∴ r² = x² + 16 [using (i)]
= (3)² + 16 = 25
⇒ r = 5cm 
∴ d = 10 cm. 

In ∆OPA, 


Similarly, 
In ∆O′PA,

= 9cm. 
∴ OO′ = OP + O′P = 5 + 9 = 14 cm.

∵ AD is the bisector of ∠BAC

∴ AD is the ⊥ bisector of BC. 
∵ O is the circumcentre of ∆ABC.
∴ AB = AC [∵ ∠ABD = ∠ACB]
(By using ∆ABD ≅ ∆ACD)

Reflex ∠AOC = 360° – (110° + 120°) 
= 130° 

∠BOC = 2 × ∠BAC [Angle subtended at center is double the angle subtended the circle]
= 2 × 30° = 60°

∠XPY = 2∠XOP
∵ ∠XOP = ∠XZY
(angles in the same segment)
∴ ∠XPY = 2∠XZY …(i)
Similarly, 
∠YPZ = 2∠YXZ …(ii)
Using (i) and (ii)
∠XPZ = 2 (∠XYZ + ∠YXZ)

∠P = ∠QPR + ∠SPR
= 64 ° + 31° 
= 95°
∵ ∠Q and ∠S are angles of the semicircle
∴ ∠Q = ∠S = 90° 
∵ PQRS is a cyclic quadrilateral
∴ ∠P + ∠R = 180°
⇒ ∠R = 180° - 95° = 85°

Reflex ∠AOC = 360° – (90° + 120°) = 150° 

In ∆ACD, 
∠ADC = 90° [∵∠ACD is angle in semicircle]
∴  AC² = DA² + DC²
⇒ (30)² = (10√5 )² + DC²
⇒ DC² = 900 – 500 
⇒ DC = √400 = 20 cm.