Find the equation of a circle passing through the points (5,7),(6,6) and (2,-2). Find its centre and radius.

Let the required equation of circle be

x² + y² + 2gx + 2fy + c = 0 ….. (1)

Since it passes through (5,7),(6,6) and (2,-2), each of these will satisfy the (1).

So,

Putting (5,7) we get,

5² + 7² + 2g(5) + 2f(7) + c = 0

⇒ 25 + 49 + 10g + 14f + c=0

⇒ 10g + 14f + c + 74 =0 …… (2)

Putting (6,6) we get,

6² + 6² + 2g(6) + 2f(6) + c = 0

⇒ 36 + 36 + 12g + 12f + c = 0

⇒12g + 12f + c + 72 = 0 …… (3)

Putting (2,-2) we get,

2² + (-2)² + 2g(2) + 2f(-2) + c = 0

⇒ 4 + 4 + 4g – 4f + c = 0

⇒ 4g – 4f + c + 8 = 0 …… (4)

Subtract (2) from (3), we get,

12g + 12f + c + 72 – (10g + 14f + c + 74) = 0

⇒ 12g + 12f + c + 72 – 10g -14 f – c – 74 = 0

⇒ 2g – 2f – 2 = 0

⇒ g – f = 1 ….. (5)

Subtract (4) from (3), we get,

12g + 12f + c + 72 – (4g – 4f + c + 8) = 0 – 0

⇒ 12g + 12f + c + 72 – 4g + 4f – c – 8 = 0

⇒ 8g + 16f + 64 = 0

⇒ g + 2f = - 8 ….. (6)

Subtract (5) from (6) to get,

g + 2f - g + f = -8 -1

⇒ 3f = -9

⇒ f = -3

Put this value in (5) to get,

g + 3 = 1

⇒ g = -2

Putting the value of g and f in (2) we get,

10(-2) + 14(-3) + c + 74 =0

⇒ -20 – 42 + c + 74 = 0

⇒ -62 + c + 74 = 0

⇒ 12 + c = 0

⇒ c = -12

Putting the values of g, f and c in (1) we get,

x² + y² + 2(-2) x + 2(-3)y + (-12) = 0

⇒ x² + y² - 4x - 6y -12 = 0

Which is required equation of circle.

Hence centre is (-g,-f) = (2,3)

Radius

= √25

= 5 units.