Question

The distance between the polar of $P(2,3)$ with respect to the circle $x^2+y^2-2x-2y+1=0$ and the polar of the inverse point of $P$ with respect to the same circle is

• A. 0
• B. $\frac{4}{\sqrt{5}}$
• C. $\frac{12}{\sqrt{5}}$
• D. 1

Answer 1

Answer: (B)

Equation of polar of point $P(2,3)$ with respect to the circle $x^2+y^2-2x-2y+1=0$ is
$T=0\Rightarrow 2x+3y-(x+2)-(y+3)+1=0$
$\Rightarrow x+2y-4=0\dots (i)$
End equation of line joining of points $P(2,3)$ and centre of the given circle $C(1,1)$ is
$y-1=\frac{2}{1}(x-1)$
$\Rightarrow 2x-y-1=0\dots (ii)$
$\because$ Inverse point of $P$ with respect to the given circle is point of intersection of polar Eq. (i) and line CP, Eq. (ii), so coordinate of the inverse point is $Q\left(\frac{6}{5},\frac{7}{5}\right)$
Now, equation of polar of point $Q\left(\frac{6}{5},\frac{7}{5}\right)$ with respect to given circle is
$\frac{6}{5}x+\frac{7}{5}y-\left(x+\frac{6}{5}\right)-\left(y+\frac{7}{5}\right)+1=0$
$\Rightarrow x+2y-8=0\dots (iii)$
$\therefore$ Required distance between polars (i) and (iii) is
$\frac{4}{\sqrt{1+4}}=\frac{4}{\sqrt{5}}$