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

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(D) Given equation of the circle is $r=12 \cos 	heta+5 \sin 	heta$.Substitute $\cos 	heta=\frac{x}{r}$ and $\sin 	heta=\frac{y}{r}$ into the equat

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$\frac{13}{2}$

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(D) Given equation of the circle is $r=12 \cos 	heta+5 \sin 	heta$.Substitute $\cos 	heta=\frac{x}{r}$ and $\sin 	heta=\frac{y}{r}$ into the equation:$r = 12 \left(\frac{x}{r}ight) + 5 \left(\frac{y}{r}ight)$$r^2 = 12x + 5y$Since $r^2 = x^2 + y^2$,we have:$x^2 + y^2 - 12x - 5y = 0$Comparing this with the general equation $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -6$ and $f = -\frac{5}{2}$.The radius of the circle is given by $\sqrt{g^2 + f^2 - c}$:Radius $= \sqrt{(-6)^2 + \left(-\frac{5}{2}ight)^2 - 0}$Radius $= \sqrt{36 + \frac{25}{4}}$Radius $= \sqrt{\frac{144 + 25}{4}} = \sqrt{\frac{169}{4}} = \frac{13}{2}$.

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

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