Question

Find the equation of the circle on the line joining the points (1, 0), (0, 1), and having its centre on the line x + y = 1.

Answer 1

Let the equation of the circle be

x² + y² + 2gx + 2fy + c = 0 ……… (1)

The circle passes through (1, 0)

1² + 0² + 2g(1) + 2f(0) + c = 0

1 + 2g + c = 0

2g + c = 1 …….. (2)

Again the circle (1) passes through (0, 1)

0² + 1² + 2g(0) + 2f(1) + c = 0

1 + 2f + c = 0

2f + c = -1 ……. (3)

(2) – (3) gives 2g – 2f = 0 (or) g – f = 0 ………. (4)

Given that the centre of the circle (-g, -f) lies on the line x + y = 1

-g – f = 1 …….. (5)

(4) + (5) gives -2f = 1 ⇒ f = `- 1/2`

Using f = `- 1/2`  in (5) we get

- g - `(- 1/2)` = 1

- g = `1 - (- 1/2) = 1/2`

g = `- 1/2`

Using g = `- 1/2` in (2) we get

`2(- 1/2) + "c" = - 1`

-1 + c = -1

c = 0

using g = `-1/2`, f = `-1/2`, c = 0 in (1) we get the equation of the circle,

`x^2 + y^2 + 2(- 1/2)x + 2(- 1/2)y + 0 = 0`

x² + y² – x – y = 0