Question

Answer the following :

Find the equation of circle passing through the point of intersection of the lines x + 3y = 0 and 2x − 7y = 0 whose centre is the point of intersection of lines x + y + 1 = 0 and x − 2y + 4 = 0

Answer 1

Let P(h, k) be the centre of the circle which is the point of intersection of the lines x + y + 1 = 0 and x − 2y + 4 = 0.

∴ h + k = −1   ...(1)

and h − 2k = −4   ...(2)

Subtracting (2) from (1), we get,

3k = 3

∴ k = 1

∴ from (1), h + 1 = −1

∴ h = − 2

∴ centre is P(− 2, 1).

Also, the circle passes through the point of intersection of the lines x + 3y = 0 and 2x − 7y = 0 which is 0 (0, 0).

∴ radius = OP = `sqrt((-2 - 0)^2 + (1 - 0)^2`

= `sqrt(4 + 1)`

= `sqrt(5)`

∴ by centre-radius form, the equation of the circle is

(x + 2)² + (y − 1)² = `(sqrt(5))^2`

∴ x² + 4x + 4 + y² − 2y + 1 = 5

∴ x² + y² + 4x - 2y = 0.