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

$S^{\text{'}}\equiv x^2+y^2-2x+2y-7=0$

$c_1(-1,-1), r_1 =\sqrt{1+1+7}=3$

Now a pair of lines $5x^2-xy-5x+y=0$ are normally to circle $S=0$

We know that normals of a circle intersect at centre 

To find the intersection point 

$P=5x^2-xy-5x+y=0$

$\frac{dP}{dx}=10x-y-5=0$

$\frac{dP}{dy}=-x+1=0$

So $x=1, y=5$

$c_2(1, 5)$

Now $c_1{c}_2=r_1+r_2$

$\Rightarrow 6=3+r_2$

$\Rightarrow r_2=3$

Now the equation of $S=0$

$(x-1{)}^2+(y-5{)}^2=9$

$\Rightarrow x^2+y^2-2x-10y+17=0$

Equation of chord of contact $AB$

$T=0$

$\Rightarrow x\left(1\right)+y(-1)-1(x+1)-5(y-1)+17=0$

$\Rightarrow x-y-x-1-5y+5+17=0$

$\Rightarrow -6y+21=0$

$\Rightarrow 2y-7=0$