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). Since the circle passing through (0,0) and (4,0).
`:.0^(2)+0^(2)+2g(0)+2f(0)+c=0:.c=0` and `4^(2)+0^(2)+2g(0)+2f(0)+0=0rArr16+8g=0` `therefore g-2`
Also, the centre (-g,-f)lies on the line y=x.
`rArr -f=-grArr f=g`
:. f=-2
now, the equation of circle having centre (2,2) and radius `sqrt8` `(x-2)^(2)+(y-2)^(2)=(sqrt8)^(2)`
`(x-2)^(2)+(y-2)^(2)=8`