A circle of radius `sqrt8 ` is passing through origin the point (4,0) . If the centre lies on the line y = x , then the equation of the circle is

Options

A. `(x-2)^(2) + (y-2)^(2) =8`
B. `(x+2)^(2) +(y +2)^(2) =8`
C. `(x-3)^(2) + (y -3)^(2) = 8`
D. `(x+3)^(2) + (y+3)^(2) =8`

Answer

A

Explanation

Let the equation of circle be ` x^(2) +y^(2) +2gx +2fy +c =0 ` , whose centre is (-g,-f) ltbr since the circle passing through (0,0) and (4,0)
` Rightarrow 0^(2) +0^(2) + 2g(0) + 2f(0)+c=0`
` Rightarrow =c=0`
` and 4^(2) +0^(2) = 2g(4) + 2f(0) + 0=0`
` Rightarrow 16 = 8 g = 0`
g=-2
f=-2
Now, the equation of circle having centre (2,2) and radius `sqrt8` is
` (x-2)^(2) + (y-2)^(2) =(sqrt8))^(2)`
` Rightarrow (x-2)^(2) + (y-2)^(2) =8`