Question

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 1

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