The locus of centre of a circle which passes through the origin and cuts off a length of 4 units on the line `x =3` is

Options

A. `y^(2)+6x=0`
B. `y^(2)+6x=13`
C. `y^(2)+6x=10`
D. `x^(2)+6y=13`

Answer

B

Explanation

Let centre of circle be C(-g,-f), then equation of circle passing through origin is `x^(2)+y^(2) +2gx+2fy=0`
`Distance, d `d=|3-(-g)|=g+3 in DeltaABC,(bc)^(2)=Ac ^(2) +BA^(2)`
` rArr g^(2)+f^(2)=(g+3)^(2)+2^(2)`
` rArr g^(2)+f^(2)=g^(2)+6g+9+4`
` therefore f^(2)=6g+13` hence, requation locus is `y^(2)+6x=13.`