Question

Find equation of circle which passes through the point $(2,3)$ and touches the line $2x-3y-13=0$ at the point $(2,-3)$.

Answer 1

$S+\lambda P=0$ where $S$ is point circle at $(2,3)$ and $P$ is tangent at $P$.
$\therefore \underset{pt.circle}{(x-2{)}^2+(y-3{)}^2}+\underset{tangent\left(By\right(n_1\left)\right)}{\lambda (x-2)}=0$
It passes through $(1,1)\therefore \lambda =5$
$\therefore x^2+y^2+x-6y+3=0$
Note: In Ist method, $S$ is given circle and in IInd method, $S$ is the point circle and hence the values of $\lambda$ are different.
Here tangent at $(1,-3)$ is $3x-y-6=0$. Hence the required family of circles is
$\underset{pt.circle}{(x-1{)}^2+(y+3{)}^2}+\underset{tangent}{\lambda (3x-y-6)=0}$
We have to choose the member of the family whose radius is $2\sqrt{10}$.
$x^2+y^2+x(3\lambda -2)+y(6-\lambda )+10-6\lambda =0.....\left(2\right)$
Its radius $=\sqrt{g^2+f^2-c}=2\sqrt{10}$
or ${\left(\frac{3\lambda -2}{2}\right)}^2+{\left(\frac{6-\lambda }{2}\right)}^2-4(10-6\lambda )=40$
or $10{\lambda }^2=160\therefore \lambda =4,-4$.
Putting in $\left(2\right)$, the required circles are
$x^2+y^2+10x+2y-14=0$
or $x^2+y^2-14x+10y+34=0$
Radius of each of the above two circles is $\sqrt{40}$ i.e. $2\sqrt{10}$.
$\therefore$ Required circle is $x^2+y^2-13=0$.