Locus of the point $R$

Question

$x+y=3$ meets $x^2+y^2-4x+6y=3$ at points $A$, $B$. A variable line meets axes at $P$, $Q$ so that $AQ$ meets $BP$ at $R$ at right angle. Find out the locus of $R$

I've started with finding points $A$, $B$ and using the condition of product of slopes $=-1$ for perpendicularity, but I didn't get the locus.

Answer 1

It is the circle having diameter $AB$

Solving the system you get $A(2,1)$ and $B(6;-3)$

Midpoint $M(4,-1)$ is the centre and $MA=\sqrt{(4-2)^2+(-1-1)^2}=\sqrt{8}$ is the radius

$(x-4)^2+(y+1)^2=8$ is the equation of the locus which expanded gives

$x^2+y^2-8 x+2 y+9=0$

Hope it helps

Answer 2

Since all triangles $ABR$ has $\angle BRA=\tfrac\pi2$, the locus of points $R$ is a circle centered at $\tfrac12(A+B)$: