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

Question

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

Dr. Abubacker · Accepted Answer

(b)
Given, $r^{2} - 8\left( \sqrt{3}\cos	heta + \sin	heta ight) + 15 = 0$
Where $r\cos	heta = x\ \ \ and\ \ y = r\sin	heta$
It can be rewritten in Cartesian form as
$x^{2} + y^{2} - (8\sqrt{3x} + y) + 15 = 0$
$\Rightarrow \ x^{2} + y^{2} - 8\sqrt{3} + 8y + 15 = 0$
Now, radius$= \sqrt{\left( 4\sqrt{3} ight)^{2} + (4)^{2} - 15} = 7$

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

Explanation

Reviewed by Dr. Abubacker

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