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)

7

Answer 2

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