Question
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.
$$\left( { - \frac{1}{2},\,\frac{1}{2}} \right)$$
B.
$$\left( {1,\,1} \right)$$
C.
$$\left( {\frac{1}{2},\,\frac{1}{2}} \right)$$
D.
none of these
Answer :
$$\left( {\frac{1}{2},\,\frac{1}{2}} \right)$$
Solution :
Let the second circle be $${x^2} + {y^2} + 2gx + 2fy = 0.$$ The common chord has the equation $$2\left( {g - 3} \right)x + 2\left( {f - 4} \right)y + 7 = 0.$$
But $$y = x$$ touches the circle.
Hence, $${x^2} + {x^2} + 2gx + 2fx = 0$$ has equal roots, i.e., $$f + g = 0.$$
$$\therefore $$ the equation of the common chord is $$2\left( {g - 3} \right)x + 2\left( { - g - 4} \right)y + 7 = 0$$
or $$\left( { - 6x - 8y + 7} \right) + g\left( {2x - 2y} \right) = 0,$$ which passes through the point of intersection of $$ - 6x - 8y + 7 = 0$$ and $$2x - 2y = 0.$$