The radius of the circle r = sqrt{3} sin thet | vedclass

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(A) Given equation of the circle is $r = \sqrt{3} \sin 	heta + \cos 	heta$.Multiplying both sides by $r$,we get $r^2 = \sqrt{3} (r \sin 	heta) + (r

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(A) Given equation of the circle is $r = \sqrt{3} \sin 	heta + \cos 	heta$.Multiplying both sides by $r$,we get $r^2 = \sqrt{3} (r \sin 	heta) + (r \cos 	heta)$.Using the polar to Cartesian conversions $x = r \cos 	heta$ and $y = r \sin 	heta$,and $r^2 = x^2 + y^2$,the equation becomes:$x^2 + y^2 = \sqrt{3} y + x$.Rearranging the terms,we get $x^2 - x + y^2 - \sqrt{3} y = 0$.Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we have $2g = -1 \Rightarrow g = -\frac{1}{2}$ and $2f = -\sqrt{3} \Rightarrow f = -\frac{\sqrt{3}}{2}$,with $c = 0$.The radius is given by $\sqrt{g^2 + f^2 - c} = \sqrt{(-\frac{1}{2})^2 + (-\frac{\sqrt{3}}{2})^2 - 0}$.Radius $= \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1$.

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

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