Find the equation

Let the required equation of circle be

x² + y² + 2gx + 2fy + c = 0 ….. (1)

As circle passes through (0,0),(-2,0)(0,3)

Putting (0,0) in (1) to get,

0+0+0+0+c = 0

⇒ c= 0

Thus (1) becomes,

x² + y² + 2gx + 2fy = 0 ….. (2)

Putting (-2,0) in (2) to get,

(-2)² + 0² – 4g + 0 = 0

⇒ 4g = 4

g = 1

Putting (0,3) in (2) to get,

0 + 3² + 0 + 6f = 0

9 + 6f = 0

6f = -9

f = -3/2

Putting g = 1, f=-3/2 in (2) we get,

x² + y² + 2x – 3y= 0