Question

Find the equation of circle passing through $(4,-3)$ and touching the lines $x+y=2$ and $x-y=2$.

Answer 1

If $(h,k)$ be the centre, then $p=r$ gives
$\frac{h+k-2}{\sqrt{2}}=\pm \frac{h-k-2}{\sqrt{2}}=\sqrt{(h-4{)}^2+(k+3{)}^2}$
Taking +ive sign, we have $k=0$
and ${\left(\frac{h-2}{\sqrt{2}}\right)}^2=(h-4{)}^2+9$
$\therefore h^2+20h+46=0$
$\therefore h=-10\pm 3\sqrt{6}$
and $r^2=(h-4{)}^2+9=\frac{(-12\pm 3\sqrt{6}{)}^2}{2}$
Taking -ive sign, we have $h=2$
$\therefore {\left(\frac{k}{12}\right)}^2=36+(k-3{)}^2$
or $k^2-12k+90=0$
Its roots are imaginary.
Hence the required circle is
$(x-h{)}^2+(y-0{)}^2=r^2$ etc.