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The radius of the circle $r=\sqrt{3} \sin \theta+\cos \theta$ is
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Verified Answer
The correct answer is:
$1$
Given equation of circle is
$$
r=\sqrt{3} \sin \theta+\cos \theta
$$
On putting $x=r \sin \theta, y=r \sin \theta$
$$
\begin{array}{lc}
\therefore & r=\sqrt{3} \frac{x}{r}+\frac{y}{r} \\
\Rightarrow & r^2=\sqrt{3} x+y \\
\Rightarrow & x^2+y^2-\sqrt{3} x-y=0 \\
\therefore \text { Radius }=\sqrt{g^2+f^2-c} \\
& =\sqrt{\left(\frac{\sqrt{3}}{2}\right)^2+\left(\frac{1}{2}\right)^2}=\sqrt{\frac{4}{4}}=1
\end{array}
$$
$$
r=\sqrt{3} \sin \theta+\cos \theta
$$
On putting $x=r \sin \theta, y=r \sin \theta$
$$
\begin{array}{lc}
\therefore & r=\sqrt{3} \frac{x}{r}+\frac{y}{r} \\
\Rightarrow & r^2=\sqrt{3} x+y \\
\Rightarrow & x^2+y^2-\sqrt{3} x-y=0 \\
\therefore \text { Radius }=\sqrt{g^2+f^2-c} \\
& =\sqrt{\left(\frac{\sqrt{3}}{2}\right)^2+\left(\frac{1}{2}\right)^2}=\sqrt{\frac{4}{4}}=1
\end{array}
$$
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