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`.
The distance between the points A and B , is

Options

A. 4
B. `4 sqrt(2)`
C. 6
D. 8

Answer

3

Explanation

We have `x^(2)+y^(2) -2x-2ay-8=0`
or `(x^(2)+y^(2)-2x-8)-2ay =0`, which is of the form `S + lambda L =0`.
Solving `S -= x^(2)+y^(2) -2x-8=0` and `L -= y =0`, we get
`x^(2)-2x-8=0`
`implies (x-4) (x+2)=0`
`implies x = - 2` or `4`
`implies A -= (4,0) , B -= (-2,0)`
Distance between A and `B =6`