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The radius of the circle with the polar equation $r^2-8 r(\sqrt{3} \cos \theta+\sin \theta)+15=0$ is
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2736 Upvotes
Verified Answer
The correct answer is:
$7$
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
$$
\begin{aligned}
& x^2+y^2-8(\sqrt{3} x+y)+15=0 \\
& \Rightarrow \quad x^2+y^2-8 \sqrt{3} x-8 y+15=0 \\
& \text { Now, radius }=\sqrt{(4 \sqrt{3})^2+(4)^2-15} \\
& =\sqrt{48+16-15}=7 \\
&
\end{aligned}
$$
$$
r^2-8 r(\sqrt{3} \cos \theta+\sin \theta)+15=0
$$
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
$$
\begin{aligned}
& x^2+y^2-8(\sqrt{3} x+y)+15=0 \\
& \Rightarrow \quad x^2+y^2-8 \sqrt{3} x-8 y+15=0 \\
& \text { Now, radius }=\sqrt{(4 \sqrt{3})^2+(4)^2-15} \\
& =\sqrt{48+16-15}=7 \\
&
\end{aligned}
$$
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