The common chord of the circle `x^2+y^2+6x+8y-7=0` and a circle passing through the origin and touching the line `y=x` always passes through the point. `(-1/2,1/2)` (b) (1, 1) `(1/2,1/2)` (d) none of these

Options

A. `(-1//2,1//2)`
B. `(1,1)`
C. `(1//2,1//2)`
D. none of these

Answer

3

Explanation

Let the second circle be `x^(2)+y^(2)+2gx+2fy=0`.
But` y=x` touches the circle.
Hence, `x^(2)+x^(2)+2gx+2fx=0` hase qual roots, i.e., `f+g=0`.
Therefore, the equation of the common chord is `2(g-3)x+ 2(-g-4)y +7 =0`
or `(-6x-8y+7) +g(2x-2y) =0`
which passes through the point of intersection of
`-6x-8y+7=0` and `2x-2y=0, i.e., (1//2,1//2)`