The squared length of the intercept made by the line `x=h` on the pair of tangents drawn from the origin to the circle `x^2+y^2+2gx+2fy+c=0` is `(4c h^2)/((g^2-c)^2)(g^2+f^2-c)` `(4c h^2)/((f^2-c)^2)(g^2+f^2-c)` `(4c h^2)/((f^2-f^2)^2)(g^2+f^2-c)` (d) none of these

Options

A. `(4ch^(2))/((g^(2)-c^(2)))(g^(2)+f^(2)-c)`
B. `(4ch^(2))/((f^(2)-c^(2)))(g^(2)+f^(2)-c)`
C. `(4ch^(2))/((g^(2)-f^(2))^(2))(g^(2)+f^(2)-c)`
D. none of these

Answer

2

Explanation

The equation of the pair of tangents from (0,0) is `S S' =T^(2)`
or `(x^(2)+y^(2)+2gx+2fy+c)c= (gx+fy+c)^(2)` (1)
The intersection points of the above pair with `x=h` are given by
`(gh+fy+c)^(2)= (h^(2)+y^(2)+2gh+2fy+c)c`
or `(f^(2)-c)y^(2)+2fghy+h^(2)(g^(2)-c)=0`

If its roots are `y_(1)` and `y_(2)`, then the length of intercept
`AB^(2)=|y_(1)-y_(2)|^(2)`
`=(y_(1)+y_(2))^(2)-4y_(1)y_(2)`
`=((2fgh)/(f^(2)-c))^(2)-4(h^(2)(g^(2)-c))/(f^(2)-c)`
`=(4ch^(2))/((f^(2)-c)^(2))(g^(2)+f^(2)-c)`