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

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(A) The 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)

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(A) The 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 coordinate transformations $x = r \cos 	heta$ 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 $2g = -1 \Rightarrow g = -1/2$ and $2f = -\sqrt{3} \Rightarrow f = -\sqrt{3}/2$,with $c = 0$.The radius of the circle is given by $\sqrt{g^2 + f^2 - c}$.Radius $= \sqrt{(-1/2)^2 + (-\sqrt{3}/2)^2 - 0} = \sqrt{1/4 + 3/4} = \sqrt{1} = 1$.

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

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