Question

The polars of $(-1,2)$ with respect to the two circles $S_1\equiv x^2+y^2+6y+7=0$ and $S_2\equiv x^2+y^2+6x+1=0$

• A. Parallel
• B. coincident
• C. Perpendicular
• D. Intersecting at a non zero point

Answer 1

Answer: (D)

$S_1\Rightarrow x^2+y^2+6y+7=0$
$S_2\Rightarrow x^2+y^2+6x+1=0$
Equation of polar of $(-1,2)$ on $S_1$ is
$-x+2y+3(y+2)+7=0$
$\Rightarrow -x+5y+13=0\dots \text{ (i) }$
Equation of polar on $S_2$ is
$\Rightarrow -x+2y+3(x-1)+1=0$
$x+y-1=0$
From Eqs. (i) and (ii)
$x=3\text{ and }y=-2$