Question

Find the equation of circle which passes through (0, 0), (2, 0), (0, 3).​

Answer 1

Let the equation of circle is x² + y² + 2gx + 2fy + c = 0

since, circle passing through (0,0) , (2,0) and (0,3) , equation of circle will satisfy all given points.

putting (0,0) in equation of circle,

(0)² + (0)² + 2g(0) + 2f(0) + c = 0

or, c = 0......(1)

putting (2,0) in equation of circle,

(2)² + (0)² + 2g(2) + 2f(0) + c = 0

4 + 4g + c = 0

from equation (1),

4 + 4g = 0 => g = -1 .......(2)

putting (0,3) in equation of circle,

(0)² + (3)² + 2g(0) + 2f(3) + c = 0

9 + 6f + c = 0

from equation (1),

9 + 6f = 0 => f = -3/2 .......(3)

now putting equations (1), (2) and (3) in equation of circle.

x² + y² + 2(-1)x + 2(-3/2)y + 0 = 0

x² + y² - 2x - 3y = 0

hence, equation of circle is x² + y² - 2x - 3y = 0