Finding the equation of a circle from the equation of its tangents

Question

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.

Answer 1

$36x^2−63xy+20y^2+54x−17y−10=0$

$\implies (12x-5y-2)(3x-4y+5)=0$

$L_{1}$: $12x-5y-2=0$

$L_{2}$: $3x-4y+5=0$

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. $

Answer 2

$0 = 36x² - 63xy + 20y² + 54x – 17y – 10 = (12x – 5y – 2)(3x – 4y + 5)$

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.