The equation of the circle,concentric with th | vedclass

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(C) The given circle is $2x^2+2y^2-6x+8y+1=0$. Dividing by $2$,we get $x^2+y^2-3x+4y+\frac{1}{2}=0$. The center is $(\frac{3}{2}, -2)$ and the radius

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$4x^2+4y^2-12x+16y-21=0$

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(C) The given circle is $2x^2+2y^2-6x+8y+1=0$. Dividing by $2$,we get $x^2+y^2-3x+4y+\frac{1}{2}=0$. The center is $(\frac{3}{2}, -2)$ and the radius $r_1$ is given by $r_1^2 = (\frac{3}{2})^2 + (-2)^2 - \frac{1}{2} = \frac{9}{4} + 4 - \frac{1}{2} = \frac{9+16-2}{4} = \frac{23}{4}$. Let the required circle be $x^2+y^2-3x+4y+k=0$. Its radius $r_2$ satisfies $r_2^2 = (\frac{3}{2})^2 + (-2)^2 - k = \frac{25}{4} - k$. Since the area is double,$\pi r_2^2 = 2(\pi r_1^2)$,so $r_2^2 = 2r_1^2$. $\frac{25}{4} - k = 2(\frac{23}{4}) = \frac{23}{2}$. $k = \frac{25}{4} - \frac{46}{4} = -\frac{21}{4}$. Substituting $k$ back,$x^2+y^2-3x+4y-\frac{21}{4}=0$. Multiplying by $4$,we get $4x^2+4y^2-12x+16y-21=0$.

The equation of the circle,concentric with the circle $2x^2+2y^2-6x+8y+1=0$ and having double its area,is

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