Consider points `A(sqrt13,0) and B(2sqrt13,0)` lying on x-axis. These points are rotated anticlockwise direction about the origin through an angle of `tan^-1(2/3)`. Let the new position of A and B be A' and B' respectively. With A' as centre and radius `2sqrt13/3` a circle `C_1` is drawn and with B' as centre and radius `sqrt13/3` circle `C_2`, is drawn. The radical axis of `C_1 and C_2` is

Options

A. `3x+2y=20`
B. `3x+2y=10`
C. `9x+6y=65`
D. none of these

Answer

3

Explanation

`theta=tan^(-1)((2)/(3))`
`:. tan theta =(2)/(3)`
`:. sin theta =(2)/(sqrt(3))` and `cos theta =(3)/(sqrt(3))`
`:. A'-= (OA cos theta , OA sin theta)`
`:. A' -= (3,2)`
Similarly, `B' -= (OB cos theta, OB sin theta ) -= (6,4)`
Now, `A'B'= r_(1)+r_(2)`
So, circles touch each other externally
The point of contact C divides `C_(1)C_(2)` internally in ratio `r_(1) : r_(2)`
`:. C-=((3(sqrt(13))/(3)-6(2sqrt(13))/(3))/(sqrt(13)/(3)+(2sqrt(13))/(3)),(2(sqrt(13))/(3)+4(2sqrt13)/(3))/((sqrt(13))/(3)+(2sqrt(13))/(3)))-= (5,(10)/(3))`
Therefore, required radical axis is a line perpendicular to `A'B'` and is passing through point C, which is
`y-(10)/(3)= - (3)/(2) (x-5)`