The radius of the circle x^2 + y^2 + 2x cos t | vedclass

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(B) The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$. Comparing the given equation $x^2 + y^2 + 2x \cos 	heta + 2y \sin 	heta - 8

Vedclass · Accepted Answer

$3$

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(B) The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$. Comparing the given equation $x^2 + y^2 + 2x \cos 	heta + 2y \sin 	heta - 8 = 0$ with the general form,we get: $2g = 2 \cos 	heta \implies g = \cos 	heta$ $2f = 2 \sin 	heta \implies f = \sin 	heta$ $c = -8$ The radius $R$ of the circle is given by the formula $R = \sqrt{g^2 + f^2 - c}$. Substituting the values: $R = \sqrt{(\cos 	heta)^2 + (\sin 	heta)^2 - (-8)}$ $R = \sqrt{\cos^2 	heta + \sin^2 	heta + 8}$ Since $\cos^2 	heta + \sin^2 	heta = 1$,we have: $R = \sqrt{1 + 8} = \sqrt{9} = 3$.

The radius of the circle $x^2 + y^2 + 2x \cos \theta + 2y \sin \theta - 8 = 0$ is

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