The radius of the circle $r=\sqrt{3} \sin heta+\cos heta$ is

Question

The radius of the circle $r=\sqrt{3} \sin 	heta+\cos 	heta$ is

Aicharya · Accepted Answer

Given equation of circle is $ r=\sqrt{3} \sin 	heta+\cos 	heta $ On putting $x=r \sin 	heta, y=r \sin 	heta$ $ \begin{array}{lc} 	herefore & 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 \ 	herefore 	ext { Radius }=\sqrt{g^2+f^2-c} \ & =\sqrt{\left(\frac{\sqrt{3}}{2}ight)^2+\left(\frac{1}{2}ight)^2}=\sqrt{\frac{4}{4}}=1 \end{array} $

The radius of the circle $r=\sqrt{3} \sin \theta+\cos \theta$ is

Solution