Question

If $3x+5y+17=0$ is polar for the circle $x^2+y^2+4x+6y+9=0$, then the pole is

• A. $(1,2)$
• B. $(-1,2)$
• C. $(2,1)$
• D. $(1,-2)$

Answer 1

The correct option is A $(1,2)$

Given circle is
$x^2+y^2+4x+6y+9=0$
Let pole be $(h,k)$
Now, the equation of polar,
$T=0\Rightarrow x\left(h\right)+y\left(k\right)+2(x+h)+3(y+k)+9=0$
$\Rightarrow (h+2)x+(k+3)y+(2h+3k+9)=0$
But, the given polar equation is
$3x+5y+17=0$
Comparing both, we get
$\frac{h+2}{3}=\frac{k+3}{5}=\frac{2h+3k+9}{17}$
Now,
$\frac{h+2}{3}=\frac{k+3}{5}\Rightarrow 5h+10=3k+9\Rightarrow h=\frac{3k-1}{5}\cdots \left(1\right)$
And
$\frac{k+3}{5}=\frac{2h+3k+9}{17}\Rightarrow 17k+51=10h+15k+45$
Using equation $\left(1\right)$, we get
$\Rightarrow 2k+6=6k-2\Rightarrow k=2\Rightarrow h=1$

Hence, the pole is $(1,2)$