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. 3
B. 6
C. `2 sqrt(3)`
D. `3 sqrt(2)`

Answer

4

Explanation

Let the tangent at points A and B interest at `P(x_(2), y_(1))`.
Then AB is chord of contact w.r.t. `P(x_(1),y_(1))`, which has equation
` x x_(1) +y y_(1) -(x+x_(1))-a(y+y_(1))-8=0`
`implies (x(x_(1)-1)+y(y_(1)-a)-(x_(1+ay_(1)+8)=0` (1)
On comparing ratio of coefficients of equations (1) and (2), we get
`(x_(1)-1)/(0)=(y_(1)-a)/(1)=(x_(1)+ay_(1)+8)/(0)= k `(say)
`implies x_(1)-1=0,y_(1)=a+k` and `x_(1)+ay_(1)+8=0`
`implies x_(1)=1` and `9+ay_(1)=0`
`:. y_(1) = - (9)/(a)`
Also, `(x_(1),y_(1))` lies on the line `x+2y+5=0`
`:. x_(1) +2y_(1)+5=0`
or `1-(18)/(a) +5=0`
`:. a=3`

So, required equation of circle `S=0` is `x^(2)+y^(2)-2x-6y-8=0`
Now, radius of circle `S=0`, is `r = sqrt (1+9+8) = 3 sqrt (2)`.