The radius of the circle with the polar equat | vedclass

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

Vedclass · Accepted Answer

$7$

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(B) The given polar equation of the circle is $r^2-8r(\sqrt{3}\cos	heta+\sin	heta)+15=0$.Using the relations $x=r\cos	heta$ and $y=r\sin	heta$,and $r^2=x^2+y^2$,we convert this to Cartesian form:$x^2+y^2-8(\sqrt{3}x+y)+15=0$$x^2+y^2-8\sqrt{3}x-8y+15=0$Comparing this with the standard form $x^2+y^2+2gx+2fy+c=0$,we get $g=-4\sqrt{3}$,$f=-4$,and $c=15$.The radius $R$ is given by $\sqrt{g^2+f^2-c}$:$R = \sqrt{(-4\sqrt{3})^2+(-4)^2-15}$$R = \sqrt{48+16-15}$$R = \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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