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

JEE MATHEMATICS MCQ: CIRCLES

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Application +4 | -1

The radius of the circle with the polar equation

{r^2} - 8r\left( {\sqrt 3 \cos \mathrm{\theta } + \sin \mathrm{\theta }} \right) + 15 = 0

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Correct option: B

Option B (7) is correct because the radius is found by converting the polar equation to Cartesian form and then calculating the radius of the resulting circle.

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