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. (a) `(-1/2,1/2)` (b) (1, 1) (c) `(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

C

Explanation

Let the equation of the circle be
`x^(2)+y^(2)+2gx+2fy+c=0 " " ...(i)`
It passes through the origin and touches the line y=x.
`:. C=0 and , |(-g+f)/(sqrt(2)|=sqrt(g^(2)+f^(2)-c)`
`rArr (f-g)^(2)=2(g^(2)+f^(2))rArr g + f = 0 rArr f = - g`
The equation of the common chord of (i) and the circle
`x^(2)+y^(2)+6x+8y-7=0` is
`2x(g-3)+2y(f-4)+c+7=0`
`rArr 2x(g-3)+2y(-g-4)+7=0 [ :' c= 0 and f=-g]`
`rArr (-6x-8y+7)+2g(x-y)=0`
It represents a line passing through the intersection of the lines -6x-8y+7=0 and x-y=0 i.e. the point (1/2, 1/2).