Question

A variable circle always touches the line $y=x$ at the origin. If all the common chords of the given circle and $x^2+y^2+6x+8y-7=0$ passes through a fixed point $(a,b)$, then $a+b$ is

Answer 1

Equation of the circle will be
$(x-0{)}^2+(y-0{)}^2+k(y-x)=0$
$\Rightarrow S_1\equiv x^2+y^2+k(y-x)=0$
$S_2\equiv x^2+y^2+6x+8y-7=0$

The common chord will be
$S_2-S_1=0$
$\Rightarrow 6x+8y-7-k(y-x)=0$
So, the line is of the form $P+\lambda Q=0$ which always passes through the point of intersection $P=0;Q=0$
$6x+8y-7=0\cdots \left(1\right)y=x\cdots \left(2\right)$
From $\left(1\right)$ and $\left(2\right)$,
$x=\frac{1}{2}=a,y=\frac{1}{2}=b$

Hence, $a+b=1$