Question

The circle $x^2+y^2-4x-4y+4=0$ is inscribed in a triangle which has two of the sides along he co-ordinate axes. The locus of the circumcentre of the triangle is $x+y-xy+k\sqrt{(x^2+y^2)}=0$, find $k$.

Answer 1

If we choose $a=2$.
The condition of tangency gives
$\frac{qa+pq-pq}{\pm \sqrt{(p^2+q^2)}}=a$,
put $a=2,p=2h$ and $q=2k$
$\left(2k\right)2+\left(2h\right)2-4hk=\pm 2\sqrt{4h^2+4k^2}$
or $h+k-hk\pm \sqrt{h^2+k^2}=0$.
Locus is $x+y-xy\pm \sqrt{x^2+y^2}=0\therefore k=\pm 1$.