Question

Find the equation of the circle which passes through the point $(1,1)$ and which touches the circles $x^2+y^2+4x-6y-3=0$ at the point $(2,3)$ on it.

Answer 1

$1st$ Method: Tangent at $(2,3)$ to given circle $S=0$ is easily found to be $P\equiv x-2=0$.
The required circle is $S+\lambda P=0$
Since it passes through $(1,1)\therefore \lambda =-3$
Hence $x^2+y^2+x-6y+3=0$
$2nd$ Method: $S+\lambda P=0$ 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.