Finding the equation of a circle from the equation of its tangents Given the equation of a pair of lines 
: $36x² - 63xy + 20y² + 54x -  17y -  10 =0.$
If the circle touches one of the lines at (-3,-1) and the other at some point then find the equation of the circle. 
I know of one way of doing this is by taking the general equation is the circle and then using $T² = SS1$ and comparing it by the given equation but it is tedious. 
So I want to know of a smarter way to do this. 
 A: $36x^2−63xy+20y^2+54x−17y−10=0$
$\implies (12x-5y-2)(3x-4y+5)=0$
$L_{1}$: $12x-5y-2=0$
$L_{2}$: $3x-4y+5=0$
Let $x^2+y^2+2ax+2by+c=0$ be the circle $C$.
Then $T=L_{1} \cap L_2=(1,2)$ and $P=C\cap L_{2}=(-3,-1)$.
Now $x(1)+y(2)+a(x+1)+b(y+2)+c=0$ is the equation of the chord (polar) $L_3$.
Substitute $(-3,-1)$ into $L_{3}$,
$(-3)(1)+(-1)(2)+a(-3+1)+b(-1+2)+c=0$
$-2a+b+c-5=0 \: \cdots \cdots (1)$
$L_{1}, L_{2}$ are equidistant from centre $O(-a,-b)$,
$\displaystyle \left| \frac{-12a+5b-2}{13} \right|=
               \left| \frac{-3a+4b+5}{5} \right|$
$5^{2}(-12a+5b-2)^{2}=13^{2}(-3a+4b+5)^{2}$
$(9a-7b-5)(7a+9b+25)=0 \: \cdots \cdots (2)$
Slope of $\displaystyle OP=\frac{-1+b}{-3+a}=-\frac{4}{3}$
$ 4a+3b=15 \: \cdots \cdots (3)$
On solving, $\displaystyle (a,b,c)=
\left( \frac{24}{11}, \frac{23}{11}, \frac{80}{11} \right) \text{ or }
\left( 14, -\frac{41}{3}, \frac{140}{3} \right)$
$C: \left \{
        \begin{array}{rcl}
          11x^{2}+11y^{2}+48x+46y+80 &=& 0 \\
          3x^{2}+3y^{2}+84x-82y+140 &=& 0
 \end{array}
\right. $
A: 
$0 = 36x² - 63xy + 20y² + 54x – 17y – 10 = (12x – 5y – 2)(3x – 4y + 5)$
Let $L_1: 3x – 4y + 5 = 0$ and $L_2: 12x – 5y – 2 = 0$
Fact-1 $P= (–3, –1)$ lies on $L_1$.
Fact-2 $L_1$ cuts $L_2$ at $T$ whose co-ordinates can be found.
Fact-3 $N$ is the normal to $L_1$ through $P$. Its equation is calculate-able.
Fact-4 $Q$, the point of intersection of $N$ and $L_2$. Its co-ordinates can be found.
Fact-5 The lengths of $TQ$ and $TP$ can be calculated.
According to the angle bisector theorem, $C$ (the center of the required circle) divides $PQ$ internally in the ratio $TP : TQ$.
According to the external angle bisector theorem, $C'$ (the center of another required circle divides $PQ$ externally in the ratio $TP : TQ$.
Result follows.
