Question

Find the equation of the circle passing through the points (5, 7), (6, 6) and (2, −2)

Answer 1

Let P (h, k) be the centre of the circle and A (5, 7), B (6, 6) and C (2, – 2) be the points on the circle.

Then PA = PB = PC

PA= PB gives: `sqrt(("h" - 5)^2 + ("k" - 7)^2`

= `sqrt(("h" - 6)^2 + ("k" - 6)^2`

On squaring both sides, we get,

h² – 10h + 25 + k² –  14k + 49

= h² –  12h + 36 + k² –  12k + 36

∴ 2h –  2k = – 2

∴ h –  k = – 1 ... (1)

PA = PC gives: `sqrt(("h" - 5)^2 + ("k" - 7)^2`

= `sqrt(("h" - 2)^2 + ("k" + 2)^2`

On squaring both sides, we get,

h² – 10h + 25 + k² – 14k + 49 = h² – 4h + 4 + k² + 4k + 4

∴ – 6h – 18k = – 66

∴ h + 3k = 11 ...(2)

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

4k = 12

∴ k = 3

∴ from (1), h – 3 = – 1

∴ h = 2

∴ the centre P is (2, 3)

∴ radius = PA 

= `sqrt((5 - 2)^2 + (7 - 3)^2`

= `sqrt(9 + 16)`

= 5

∴ the equation of the required circle is

(x – 2)² + (y – 3)² = 5²

∴ x² – 4x + 4 + y² – 6y + 9 = 25

∴ x² + y² – 4x – 6y – 12 = 0.