A circle of area 9π square units has two of its diameters along the l

1.	A circle of area 9π square units has two of its diameters along the lines x + y = 5 and x – y = 1. Find the equation of the circle.
Answer» Area of the circle = 9π (i.e) πr² = 9π ⇒ r² = 9 ⇒ r = 3 (i.e) radius of the circle = r = 3 The two diameters are x + y = 5 and x – y = 1 The point of intersection of the diameter is the centre of the circle = C To find C: Solving x + y = 5 … (1) x – y = 1 …(2) (1) + (2) ⇒ 2x = 6 ⇒ x = 3 Substituting x = 3 in (1) we get 3 + y = 5 ⇒ y = 5 – 3 = 2 ∴ Centre = (3, 2) and radius = 3 So equation of the circle is (x – 3)² + (y – 2)² = 3² (i.e) x² + y² – 6x – 4y + 4 =0

A circle of area 9π square units has two of its diameters along the lines x + y = 5 and x – y = 1. Find the equation of the circle.

Answer»

Area of the circle = 9π

(i.e) πr² = 9π

⇒ r² = 9 ⇒ r = 3

(i.e) radius of the circle = r = 3

The two diameters are x + y = 5 and x – y = 1

The point of intersection of the diameter is the centre of the circle = C

To find C:

Solving x + y = 5 … (1)

x – y = 1 …(2)

(1) + (2) ⇒ 2x = 6 ⇒ x = 3

Substituting x = 3 in (1) we get

3 + y = 5 ⇒ y = 5 – 3 = 2

∴ Centre = (3, 2) and radius = 3

So equation of the circle is

(x – 3)² + (y – 2)² = 3²

(i.e) x² + y² – 6x – 4y + 4 =0