If a variable circle S=0 touches the line y = x and passes through the point (0, 0), then the fixed point that lies on the common chord of the circles $x^2+y^2+6x+8y-7=0$and S=0 is

$$\begin{gathered} For the given conditions equation of circle is \\ {(x-0)}^2+{(y-0)}^2+\lambda (x-y)=0 \\ \Rightarrow S\equiv x^2+y^2+\lambda x-\lambda y=0\to \left(1\right) \\ Given circle{x}^2+y^2+6x+8y-7=0\to \left(2\right) \\ common chord of \left(1\right) and \left(2\right) is (\lambda -6)x-(\lambda +8)y+7=0\to \left(3\right) \\ By verification first option the point (\frac{1}{2},\frac{1}{2}) lies on \left(3\right) \end{gathered}$$

$$\begin{gathered} {\mathrm{g}}^2+{\mathrm{f}}^2=\frac{{\mathrm{f}}^2+{\mathrm{g}}^2-2\mathrm{gf}}{2} \\ 2{\mathrm{g}}^2+2{\mathrm{f}}^2={\mathrm{f}}^2+{\mathrm{g}}^2-2\mathrm{gf} \\ {\mathrm{g}}^2+{\mathrm{f}}^2+2\mathrm{gf}=0 \\ {(\mathrm{g}+\mathrm{f})}^2=0 \\ \mathrm{g}+\mathrm{f}=0 \end{gathered}$$

$$\begin{gathered} \mathrm{common} \mathrm{chord} \mathrm{S}-{\mathrm{S}}^1=0 \\ 2\mathrm{gx}-6\mathrm{x}+2\mathrm{fy}-8\mathrm{y}+7=0 \\ 2\mathrm{gx}-6\mathrm{x}+2\mathrm{fx}-8\mathrm{x}+7=0 \\ 2\mathrm{gx}-14\mathrm{x}-2\mathrm{gx}+7=0 \\ \mathrm{x}=\frac{1}{2} \mathrm{y}=\frac{1}{2} \end{gathered}$$