Question

A triangle has two of its sides along the axes, its third side touches the circle $x^2+y^2-2ax-2ay+a^2=0$.Prove that the locus of the circum-centre of the triangle is $a^2-2a(x+y)+2xy=0$.

Answer 1

The given circle has its centre $(a,a)$ and radius $a$ so that this circle touches both the axes along which lie the two sides of the triangles.
Let the third side be $x/p+y/q=1$,
so that $P$ is $(p,0)$ and $Q$ is $(0,q)$. The line $PQ$ touches the given circle at the point $R$.
Since $\angle POQ$ is a right angle therefore $PQ$ is diameter of the circle passing through the points $O,P$ and $Q$ and its centre is mid-point $R$ of $PQ$.
$\therefore R$ is $(p/2,q/2)=(h,k)$, say.
$p=2h,q=2k$
Now the line $PQ$ is $x/p+y/q=1$
or $qx+py-pq=0$
touches the given circle. $\therefore p=r$
or $\frac{qa+pa-pq}{\sqrt{(p^2+q^2)}}=a$. Square
$a^2(p+q{)}^2+p^2{q}^2-2pqa(p+q)=a^2(p^2+q^2)$
or $a^2.2pq+p^2{q}^2-2pqa(p+q)=0$ cancel $pq$
or $2a^2-2a(p+q)+pq=0$
or $2a^2-2a(2h+2k)+2h.2k=0$
$\therefore$ Locus of centre $(h,k)$ is
$a^2-2a(x+y)+2xy=0$.