Question

if a point p x, y lies on a circle whose centre is 3 - 2 and radius is equal to 3 show that x square + y square - 6 x + 4 y + 4 is equal to zero

Answer 1

Equation of circle when centre (g,f) and radius a is given :

(x - g)2 + (y - f)2 = a2

Where g = 3,f = -2, a = 3

Putting values in equation, we have

(x - 3)2 + (y + 2)2 = 32

x2- 6x + 9 + y2 + 4y + 4 = 9

x2 + y2 - 6x + 4y + 13 = 9

x2 + y2- 6x + 4y + 13 - 9 = 0

x2 + y2 - 6x + 4y + 4 = 0

Hence proved