Question

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

• A. 8
• B. 7
• C. 6
• D. 5

Answer 1

Answer: (B)

Given polar equation of circle is
$r^{2}-8 r(\sqrt{3} \cos \theta+\sin \theta)+15=0$
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-8 y+15=0 $
Now, radius $=\sqrt{(4 \sqrt{3})^{2}+(4)^{2}-15} $
$=\sqrt{48+16-15}=7$