Question

The line $y=x$ touches a circle at $P$ so that $OP=4\sqrt{2}$, where $O$ is the origin. The point $(-10,2)$ is inside the circle and the length of the chord on the line $x+y=0$ is $6\sqrt{2}$. Find the equation of the circle.

Answer 1

$OPCL$ is a rectangle in which $OP=4\sqrt{2}=CL$. Also $RQ=6\sqrt{2}$. Hence
$r^2=(4\sqrt{2}{)}^2+(3\sqrt{2}{)}^2=50$
If the centre be $(h,k)$, then the circle is
$(x-h{)}^2+(y-k{)}^2=50.....\left(1\right)$
$y-x=0$ is a tangent $\therefore p=r$
or $\frac{k-h}{\pm \sqrt{2}}=5\sqrt{2}\therefore k-h=\pm 10....\left(2\right)$
The distance of $C(h,k)$ from $y+x=0$ is $4\sqrt{2}$.
$\therefore \frac{k+h}{\pm \sqrt{2}}=4\sqrt{2}\therefore k+h=\pm 8....\left(3\right)$
Eq. $\left(2\right)$ and $\left(3\right)$ will make four sets of equations which when solved will give the centres as $(9,-1),(1,-9),(-1,9),(-9,1)$.
Hence the equations of all the possible circles are
$(x-9{)}^2+(y+1{)}^2-50=0.....\left(1\right)$
$(x-1{)}^2+(y+9{)}^2-50=0.....\left(2\right)$
$(x+1{)}^2+(y-g{)}^2-50=0....\left(3\right)$
and $(x+9{)}^2+(y-1{)}^2-50=0....\left(4\right)$
Since the point $(-10,2)$ lies inside the circle, its co-ordinates must make the left hand side of the required circle negative. This condition is satisfied by $\left(4\right)$ only. Hence the equation of the required circle is $x^2+y^2+18x-2y+32=0$.