The lines represented by $5x^2-xy-5x+y=0$are normals to a circle S = 0. If this circle touches the circles $S\text{'}\equiv x^2+y^2-2x+2y-7=0$ externally, then the equation of the chord of contact of centre of $S\text{'}=0$ with respect to S = 0 is

$$\begin{gathered} Given pair of straight line 5{\mathrm{x}}^2-\mathrm{xy}-5\mathrm{x}+\mathrm{y}=0 which is normal to the circle S\equiv 0 \\ \mathrm{x}\left(5\mathrm{x}-\mathrm{y}\right)-1\left(5\mathrm{x}-\mathrm{y}\right)=0 \\ \Rightarrow \left(\mathrm{x}-1\right)( 5\mathrm{x}-\mathrm{y})=0 \\ \Rightarrow (x-1)=0 or 5x-y=0 \\ \Rightarrow x=1 and \mathrm{y}=5 \\ So, {\mathrm{C}}_1=(1, 5) , l\mathrm{et} {\mathrm{r}}_1=\mathrm{r} \\ \mathrm{For} \mathrm{the} \mathrm{given} \mathrm{circle} \\ {\mathrm{C}}_2=(1, -1), {\mathrm{r}}_2=3 \\ Now {\mathrm{C}}_1 {\mathrm{C}}_2={\mathrm{r}}_1+{\mathrm{r}}_2 \\ \Rightarrow 6=\mathrm{r}+3 \\ \Rightarrow \overline{)\mathrm{r}=3} \end{gathered}$$

Equation  of circle S=0 is

$$\begin{gathered} {(\mathrm{x}-1)}^2+{(\mathrm{y}-5)}^2=3^2 \\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}-10\mathrm{y}+17=0 \end{gathered}$$

The chord of contact of (1, -1) w:r $\mathrm{S}=0 \mathrm{is} {\mathrm{S}}_1=0$

$$\begin{gathered} \Rightarrow 1\left(\mathrm{x}\right)+(-1)\mathrm{y}-(\mathrm{x}+1)-5(\mathrm{y}-1)+17=0 \\ \Rightarrow \mathrm{x}-\mathrm{y}-\mathrm{x}-1-5\mathrm{y}+5+17=0 \\ \Rightarrow -6\mathrm{y}+21=0 \\ \Rightarrow 6\mathrm{y}-21=0 \\ \Rightarrow 2\mathrm{y}-7=0 \end{gathered}$$