The equation of the concentric circle,with th | vedclass

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(D) Given equation of circle $C_1$: $x^2+y^2-6x-4y-12=0$.Rewriting in standard form: $(x^2-6x+9) + (y^2-4y+4) = 12+9+4$.$(x-3)^2 + (y-2)^2 = 25$.Thus,

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$x^2+y^2-6x-4y=37$

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(D) Given equation of circle $C_1$: $x^2+y^2-6x-4y-12=0$.Rewriting in standard form: $(x^2-6x+9) + (y^2-4y+4) = 12+9+4$.$(x-3)^2 + (y-2)^2 = 25$.Thus,the center is $(3,2)$ and the radius $r = 5$.Area of $C_1 = \pi r^2 = 25\pi$.Let the radius of the required concentric circle be $R$.Given that the area of the required circle is double the area of $C_1$:$\pi R^2 = 2 	imes (25\pi) = 50\pi$.$R^2 = 50$.The equation of the concentric circle with center $(3,2)$ is $(x-3)^2 + (y-2)^2 = R^2$.$(x-3)^2 + (y-2)^2 = 50$.$x^2-6x+9 + y^2-4y+4 = 50$.$x^2+y^2-6x-4y+13 = 50$.$x^2+y^2-6x-4y = 37$.

The equation of the concentric circle,with the circle $C_1$ having equation $x^2+y^2-6x-4y-12=0$ and having double the area compared to the area of $C_1$,is

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