PQ subtends an angle of $45^{\circ}$ to the origin. Find the locus of the mid point of a chord PQ to circle $x^2+y^2-2ax-2by=0$

Question

A chord PQ of the circle $x^2+y^2-2ax-2by=0$ subtends an angle of $45^{\circ}$ to the origin.

Given that M is the mid point of PQ. Find the equation of the locus of M.

MY ATTEMPT Centre of the given circle $= \frac{-1}{2}[(-2a),(-2b)]=C(a,b)$

Let, $M(x,y)$

Slope of $CM$=$(\frac{b-y}{a-x})$

$CM\perp PQ$

$\therefore$ Slope of $PQ$=-$(\frac{a-x}{b-y})$

Equation of $PQ$ gives locus of $M(x,y)$

$\implies$$y_{PQ}=m_{PQ}x+c$

$\implies$$y=-(\frac{a-x}{b-y})+c$

But my approach above is wrong since the textbook answer is as below:

$x^2+y^2-2ax-2by+\frac{1}{2}(a^2+b^2)=0$

YOUR HELP WILL BE HIGHLY APPRECIATED

Answer 1

The given circle of center $O(a,b)$ and radius $r=\sqrt{a^2+b^2}$ contains the origin. Then, the chord $PQ$ subtends a right-angle to $O$ and the distance from $O$ to $PQ$ is $\frac1{\sqrt2}r$.

Thus, the locus is on a circle of center $(a,b)$ and radius $\frac1{\sqrt2}r$, which determines its equation as $$(x-a)^2+(y-b)^2=\frac{1}{2}(a^2+b^2)$$

Answer 2

Hint: $O$ the origin is on the circle, $\angle POQ = 45^\circ$ means $\angle PCQ = 90^\circ$.

$M$ is on a circular arc centred at $C$ with shorter radius.