Question

Polar of the circle $x^2+y^2=5$ with respect to the point $(1,-2)$ is a line for the circle $x^2+y^2-8x+6y+20=0$ which is

• A. a tangent
• B. cut at two real points
• C. cut at two imaginary points
• D. None of the above

Answer 1

Answer: (A)

Equation of polar of the circle
$x^2+y^2=5$
with respect to the point $(1,-2)$ is $x(1)+y(-2)=5$
$\Rightarrow$ $x-2y=5$
$\Rightarrow$ $x-2y-5=0$
Centre of the circle $x^2\text{+}{y}^2-8x+6y+20=0$ is $(4,-3)$ and radius $=\sqrt{16+9-20}=\sqrt{5}$
Perpendicular distance from the centre
$(4,-3)$ to the line $x-2y-5=0$
$=\left|\frac{4+6-5}{\sqrt{1+4}}\right|$
$=\left|\frac{5}{\sqrt{5}}\right|=\sqrt{5}=$ radius of circle
Hence, this is the tangent to the circle.