Consider a circle C_1: x^2+y^2-4x-2y=alpha-5. | vedclass

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(B) The equation of circle $C_1$ is $x^2+y^2-4x-2y+5-\alpha=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,the center is $(2, 1)$ and the radius $r_1 = \sqr

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$2$

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(B) The equation of circle $C_1$ is $x^2+y^2-4x-2y+5-\alpha=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,the center is $(2, 1)$ and the radius $r_1 = \sqrt{2^2+1^2-(5-\alpha)} = \sqrt{\alpha}$. The line of reflection is $2x-y+1=0$. Let the center of $C_2$ be $(f, g)$. The image of $(2, 1)$ in $2x-y+1=0$ is given by $\frac{f-2}{2} = \frac{g-1}{-1} = \frac{-2(2(2)-1+1)}{2^2+(-1)^2} = \frac{-2(4)}{5} = -\frac{8}{5}$. Thus,$f = 2 - \frac{16}{5} = -\frac{6}{5}$ and $g = 1 + \frac{8}{5} = \frac{13}{5}$. The equation of $C_2$ is $x^2+y^2-2fx-2gy+\frac{36}{5}=0$. The radius $r$ of $C_2$ is $\sqrt{f^2+g^2-\frac{36}{5}} = \sqrt{\frac{36}{25}+\frac{169}{25}-\frac{180}{25}} = \sqrt{\frac{25}{25}} = 1$. Since reflection preserves the radius,$r = r_1 = \sqrt{\alpha} = 1$,so $\alpha = 1$. Therefore,$\alpha+r = 1+1 = 2$.

Consider a circle $C_1: x^2+y^2-4x-2y=\alpha-5$. Let its mirror image in the line $y=2x+1$ be another circle $C_2: 5x^2+5y^2-10fx-10gy+36=0$. Let $r$ be the radius of $C_2$. Then $\alpha+r$ is equal to $......$.

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