Class 9 Maths - Circles Previous Year Questions

Very Short Answer Type Questions 

Q1. If points A, B and C are such that AB ⊥ BC and AB = 12 cm, BC = 16 cm. Find the radius of the circle passing through the points A, B and C.
 Sol: 

∵ AB ⊥ BC
∴ ∠B = 90°                  [Angle in a semicircle]
⇒ AC is a diameter 

AC = 20

Radius = 10cm

Q2. The angles subtended by a chord at any two points of a circle are equal. Write true or false for the above statement and justify your answer.
Sol: 

False. If two points lie in the same segment only, then the angles will be equal otherwise they are not equal.

Short Answer Type Questions

Q1. In the figure AOC is a diameter of the circle and arc AXB = (1/2) arc BYC. Find ∠BOC.
Sol:

∵ arc AXB = (1/2)arc BYC
∴ ∠AOB = (1/2) ∠BOC

Also ∠AOB + ∠BOC = 180°
⇒  (1/2)∠BOC + ∠BOC = 180°
⇒  (3/2) ∠BOC = 180°
⇒ ∠BOC =  (1/2) × 180° = 120°

Q2. In the figure ∠ABC = 45°. Prove that OA ⊥ OC.
Sol: 

Since the angle subtended at the centre by an arc is double the angle subtended by it at any other point on the remaining part of the circle.

∴ ∠ABC =(1/2)∠AOC
⇒ ∠AOC = 2 ∠ABC = 2 × 45° = 90°                  [∵ ∠ABC = 45°]
Thus, OA ⊥ OC.

Q3. In the adjoining figure, O is the centre of the circle. Find the length of AB.
Sol:

Since chord AB and chord CD subtend equal angles at the centre,
i.e. ∠ AOB = ∠ COD                  [Each = 60°]

∴ Chord AB = Chord CD
⇒ Chord AB = 5 cm                  [∵ Chord CD = 5 cm]
Thus, the required length of chord AB is 5 cm.

Q4. In the adjoining figure, O is the centre of the circle and OP = OQ. If AP = 4 cm, then find the length of CD.
Sol:

 ∵ OP = OQ
∴ Chord AB and chord CD are equidistant from the centre.

Thus, the required length of CD is 8 cm.

Q5. If O is the centre of the circle, then find the value of x.
 Sol:

 ∵ AOB is a diameter.
∴ ∠ AOC + ∠ COB = 180°                  [Linear pairs]
⇒ 130° + ∠ COB = 180°
⇒ ∠ COB = 180° - 130° = 50°
Now, the arc CB is subtending ∠COB at the centre and ∠CDB at the remaining part.

∴ ∠ CDB = (1/2)∠COB
⇒ ∠ CDB = (1/2)x 50° = 25°
Thus, the measure of x = 25°

Q6. The radius of a circle is 17 cm. A chord of length 30 cm is drawn. Find the distance of the chord from the centre.
Sol:

Length of chord AB = 30 cm.
Since, OP ⊥ AB
∴ P is the mid-point of AB.

AP = (1/2)AB =(1/2) x 30 cm = 15 cm
Now, in right ΔAPO, AO² = AP² + OP²
⇒ 17² = 15² + OP²
∴ OP² = 17² - 15² = (17 - 15)(17 + 15)
= 2 x 32 = 64
⇒ OP = 64 = 8 cm
∴ The distance of the chord AB from the centre O is 8 cm.

Long Answer Type Questions

Q1. An equilateral triangle is inscribed in a circle. Find the radius of the circle.
Sol: 

Let ABC be an equilateral triangle such that AB = BC = AC = 9 cm                  (each)
Let us draw a median AD corresponding to BC.
∴ BD =(1/2) BC
⇒ BD = (1/2) x 9 cm = (9/2)cm
Also, AD ⊥ BC                  [∵ O is the centre of the circle]
Now, in right ΔADB,

AD² = AB² - BD²

Since, in an equilateral triangle, the centroid and circumcentre coincide.

∴ AO: OD = 2:1

⇒  

⇒ Radius = 3√3 cm

Thus, the required radius =  3√3 cm