The equation of the locus of the middle point of a chord of the circle

$x^2+y^2-2x-2y=0$ such that the pair of lines joining the origin to the point of intersection of the chord and the circle are equally inclined to the x-axis is

Let $(h, k)$ be the mid-point of a chord of the circle

$x^2+y^2-2x-2y=0$ Then, the equation of the chord is

$\begin{array}{l}hx+ky-(x+h)-(y+k)=h^2+k^2-2h-2k\\ \Rightarrow x(h-1)+y(k-1)=h^2+k^2-h-k\end{array}$

The combined equation of the straight lines joining the origin to the points of intersection of the circle and the chord (i) is

$x^2+y^2-2(x+y)\left\{\frac{x(h-1)+y(k-1)}{h^2+k^2-h-k}\right\}=0$

$\begin{array}{l}\Rightarrow x^2\left(h^2+k^2-3h-k+2\right)+y^2\left(h^2+k^2\right.-h-3k+2)\\ -2xy(h+k-2)=0\end{array}$

Lines represented by this equation are equally inclined to the x-axis.

$\therefore$ Sum of their slopes $= 0$

$\Rightarrow \frac{2(h+k-2)}{h^2+k^2-h-3k+2}=0\Rightarrow h+k-2=0\Rightarrow h+k=2$

Hence, the locus of $(h, k)$ is$x + y = 2.$