Difficult coordinate geometry problem The base of a triangle passes through a fixed point $P(a,b)$ and its sides are respectively bisected at right angles by the lines $x+y=0$ and $x=9y$. If the locus of the third vertex is a circle, then find its equation.
Apart from the fact that the circumcentre of the given triangle is the origin $(0,0)$ , I havent been able to find anything else. Also since the geometry given in the problem is fixed, its a little difficult to plot. Any hints or clues??
Thanks in advance!!...
The correct answer given in the key is :
$4x^2 + 4y^2 + (5a+4b)x + (4a-5b)y=0$
 A: Let $A(x,y)$ be the required locus.
First reflect $A(x,y)$ about $x+y=0$, we get
$$B=(-y,-x)$$
Now reflect $A(x,y)$ about $x-9y=0$, we get $C=(x',y')$
$$\frac{y-y'}{x-x'}=-9 \tag{1}$$
Also,
$$\frac{x-9y}{\sqrt{1+9^2}}=-\frac{x'-9y'}{\sqrt{1+9^2}}$$
$$x+x'=9(y+y') \tag{2}$$
Solving $(1)$ and $(2)$,
$$C(x',y')=\left( \frac{40x+9y}{41},\frac{9x-40y}{41} \right)$$
Now $BC$ contains $P$, then
\begin{align*}
  \frac{-x-b}{-y-a} &= \frac{-x-\dfrac{9x-40y}{41}}{-y-\dfrac{40x+9y}{41}}
  \\[5pt]
  \frac{x+b}{y+a} &= \frac{50x-40y}{40x+50y} \\[5pt]
  (x+b)(4x+5y) &= (y+a)(5x-4y)
\end{align*}
Hence, the locus is
$$\fbox{$4x^2+4y^2-(5a-4b)x+(4a+5b)y = 0$} \\$$

A: Well, let's call $(x,y)$ the third point and $(m,n)$ and $(p,q)$ the base points.
The middle point bettwen $(x,y)$ and $(p,q)$ belongs to $x=9y$ so 
$$x+p=9(y+q) \quad (1)$$
For the same reason, but now for $(x,y)$ and $(m,n)$
$$x+m+y+n=0 \quad (2)$$
Once $(0,0)$ is the circuncenter
$$x^2+y^2=p^2+q^2=m^2+n^2 \quad (3)$$
Now, solving the system $x+p=9(y+q)$ and $x^2+y^2=p^2+q^2$ we get
$$41x=9q+40p$$
$$41y=9p-40q$$
Solving the system $x+m=-(y+n)$ and $x^2+y^2=m^2+n^2$ we get:
$$n=-x$$
$$m=-y$$
Finaly once $P=(a,b)$ belongs to the line $(p,q)$ and $(m,n)$ then
$$\frac{n-q}{m-p}=\frac{q-b}{p-a}$$
And replacing $m,n,p,q$ as a function of $x,y$ we get
$$4x^2 + 4y^2 + (5a+4b)x + (4a-5b)y=0$$
