Question

Find the pole of the line $3x+5y+17=0$ with respect to the circle $x^{{}^2}+y^2+4x+6y+9=0$.

Answer 1

Given equation of circle:-

$x^2+y^2+4x+6y+9=0$

Given equation of line:-

$3x+5y+17=0.....\left(1\right)$

Let $P(a,b)$ be the pole of line $3x+5y+17=0$ w.r.t. the circle $x^2+y^2+4x+6y+9=0$

Now equation of polar of point $P(a,b)$ w.r.t. the circle $x^2+y^2+4x+6y+9=0$-

$ax+by+2(x+a)+3(y+b)+9=0$

$\Rightarrow (a+2)x+(b+3)y+(2a+3b+9)=0.....\left(2\right)$

As the equation $\left(2\right)\&\left(3\right)$ are same, therefore,

$\frac{a+2}{3}=\frac{b+3}{5}=\frac{2a+3b+9}{17}$

Now,

Case I:-

$\frac{a+2}{3}=\frac{b+3}{5}$

$\Rightarrow 5(a+2)=3(b+3)$

$\Rightarrow 5a+10=3b+9$

$\Rightarrow 3b-5a=1.....\left(3\right)$

Case II:-

$\frac{a+2}{3}=\frac{2a+3b+9}{17}$

$17(a+2)=3(2a+3b+9)$

$17a+34=6a+9b+27$

$9b-11a=7.....\left(4\right)$

Multiplying ${eq}^n\left(3\right)$ by $3$, we get

$9b-15a=3.....\left(5\right)$

Subtracting ${eq}^n\left(5\right)$ from $\left(4\right)$, we have

$(9b-11a)-(9b-15a)=7-3$

$9b-11a-9b+15a=4$

$4a=4$

$\Rightarrow a=1$

Substituting the value of $a$ in ${eq}^n\left(3\right)$, we have

$3b-5\left(1\right)=1$

$\Rightarrow 3b=1+5$

$\Rightarrow b=2$

Hence the point $P$ is $(1,2)$.

Hence the the pole of the line $3x+5y+17=0$ w.r.t. the circle $x^2+y^2+4x+6y+9=0$ is $(1,2)$.