The radius of the circle r = 12 cos theta + 5 | vedclass

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(D) Given equation of the circle is $r = 12 \cos 	heta + 5 \sin 	heta$.Using the substitutions $x = r \cos 	heta$ and $y = r \sin 	heta$,we multip

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$\frac{13}{2}$

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(D) Given equation of the circle is $r = 12 \cos 	heta + 5 \sin 	heta$.Using the substitutions $x = r \cos 	heta$ and $y = r \sin 	heta$,we multiply the equation by $r$ to get $r^2 = 12(r \cos 	heta) + 5(r \sin 	heta)$.Substituting $r^2 = x^2 + y^2$,we get $x^2 + y^2 = 12x + 5y$.Rearranging the terms,we have $x^2 - 12x + y^2 - 5y = 0$.Completing the square for $x$ and $y$: $(x - 6)^2 - 36 + (y - \frac{5}{2})^2 - \frac{25}{4} = 0$.This simplifies to $(x - 6)^2 + (y - \frac{5}{2})^2 = 36 + \frac{25}{4} = \frac{144 + 25}{4} = \frac{169}{4}$.Comparing this with the standard form $(x - h)^2 + (y - k)^2 = R^2$,the radius $R = \sqrt{\frac{169}{4}} = \frac{13}{2}$.

The radius of the circle $r = 12 \cos \theta + 5 \sin \theta$ is

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