Consider the family of circles `x^(2)+y^(2)-2x-2ay-8=0` passing through two fixed points A and B . Also, `S=0` is a cricle of this family, the tangent to which at A and B intersect on the line `x+2y+5=0`.
If the circle `x^(2)+y^(2)-10x+2y=c=0` is orthogonal to `S=0`, then the value of c is

Options

A. 8
B. 9
C. 10
D. 12

Answer

4

Explanation

Circle `x^(2)+y^(2)-10x+2y+c=0` is orthogonal to `x^(2)+y^(2)-2x-6y-8=0`.
So, by applying condition of orthogonal intersection, we get
`2(-5)(-1)+2(1) (-3)= c-8`
`:. C= 12`