The radius of the circle with the polar equat | vedclass

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(B) The given polar equation of the circle is $r^2-8r(\sqrt{3} \cos 	heta + \sin 	heta) + 15 = 0$. Substituting $r \cos 	heta = x$ and $r \sin 	he

Vedclass · Accepted Answer

$7$

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(B) The given polar equation of the circle is $r^2-8r(\sqrt{3} \cos 	heta + \sin 	heta) + 15 = 0$. Substituting $r \cos 	heta = x$ and $r \sin 	heta = y$,and noting that $r^2 = x^2 + y^2$,the equation becomes: $x^2 + y^2 - 8(\sqrt{3}x + y) + 15 = 0$ $x^2 + y^2 - 8\sqrt{3}x - 8y + 15 = 0$. Comparing this with the general equation of a circle $x^2 + y^2 + 2gx + 2fy + c = 0$,we have $g = -4\sqrt{3}$,$f = -4$,and $c = 15$. The radius $R$ is given by $\sqrt{g^2 + f^2 - c}$. $R = \sqrt{(-4\sqrt{3})^2 + (-4)^2 - 15} = \sqrt{48 + 16 - 15} = \sqrt{49} = 7$.

The radius of the circle with the polar equation $r^2-8r(\sqrt{3} \cos \theta + \sin \theta) + 15 = 0$ is

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