Question

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

Answer 1

Let the circle be $(x-h{)}^2+(y-k{)}^2=r^2$.
Centre lies on either of bisectors of angle between the tangents which are
$\frac{x+y-2}{\sqrt{2}}=\pm \frac{x-y-2}{\sqrt{2}}$ or $y=0$ or $x=2$
$\therefore k=0$ or $h=2$
Again by applying $p=r$ with any tangent
$\frac{h+k-2}{\sqrt{2}}=r$ Put $k=0\therefore (h-2{)}^2=2r^2.....(1)$
Again the circle passes through $(-4,3)$.
$\therefore (4+h{)}^2+(3-k{)}^2=r^2$ Put $k=0$
$h^2+8h+25=\frac{(h-2{)}^2}{2}$
or $h^2+20h+46=0\therefore h=-10\pm 3\sqrt{6}$
and $r^2=\frac{1}{2}(h-2{)}^2=\frac{1}{2}(-12\pm 3\sqrt{16}{)}^2=99\pm 36\sqrt{6}$
$\therefore$ Circle is $(x+10\pm 3\sqrt{6}{)}^2+y^2=99\pm 36\sqrt{6}$
Note: If we take the other bisector, then $h=2$ and as above the quadratic in $k$ will be $k^2-12k+96=0$. Its roots are imaginary.