Suppose `a x+b y+c=0` , where `a ,ba n dc` are in `A P` be normal to a family of circles. The equation of the circle of the family intersecting the circle `x^2+y^2-4x-4y-1=0` orthogonally is `x^2+y^2-2x+4y-3=0` `x^2+y^2+2x-4y-3=0` `x^2+y^2-2x+4y-5=0` `x^2+y^2-2x-4y+3=0`

Options

A. `x^(2)+y^(2)-2x+4y-3=0`
B. `x^(2)+y^(2)+2x-4y-3=0`
C. `x^(2)+y^(2)-2x+4y-5=0`
D. `x^(2)+y^(2)-2x-4y+3=0`

Answer

1

Explanation

a,b, and c are in AP . So, `ax+by=c=0` represents a family of lines passing through the point (1,-2). So, the family of circles (concentric ) will be given by `x^(2)+y^(2)-2x+4y+c=0`. It intersects the given circle orthogonally. Therefore,
`2(-1)(-2)+2(2)(-2)=-1+c` or `c= -3`