JEE Class main Answered

if the point (h, k) is center of circle and r is radius of circle , then equation of circle is

(x-h)² + (y-k)² = r²  .................................(1)

if the centre of circle is in x-axis , then k =  0

If the circle passes through (-2,3), then we get from eqn.(1)

(-2-h)² + 9 = r²  ...........................(2)

If the circle passes through ( 4, 5), then we get from eqn.(2)

(4 -h )² + 25 = r²  ............................ (3)

By equating eqn.(2) and (3) , we get

(4 -h )² + 25 = (h+2)² + 9

By solving above expression, we get h = 7/3

By substituting h = 7/3 in eqn.(1) , we get r²= ( 250 / 9 )

Hence equation of circle is

[ x - (7/3) ]² + y² = (250/9)

(3x-7)² + 9y² = 250

9x² + 9 y² - 42x -201 = 0