Question

The radius of the circle with the polar equation $r^2-8r(\sqrt{3}\cos \theta +\sin \theta )+15=0$ is

• A. 8
• B. 7
• C. 6
• D. 5

Answer 1

Answer: (B)

Given polar equation of circle is
$r^2-8r(\sqrt{3}\cos \theta +\sin \theta )+15=0$
or $r^2-8(\sqrt{3}r\cos \theta +r\sin \theta )+15=0$
where $r\cos \theta =x$ and $y=r\sin \theta$.
It can be rewritten in cartesian form
$x^2+y^2-8(\sqrt{3}x+y)+15=0$
$\Rightarrow x^2+y^2-8\sqrt{3}x-8y+15=0$
Now, radius $=\sqrt{(4\sqrt{3}{)}^2+(4{)}^2-15}$
$=\sqrt{48+16-15}=7$