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.