Let the mirror image of a circle c_{1}: x^{2} | vedclass

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(C) The circle $c_{1}$ is $x^{2}+y^{2}-2x-6y+\alpha=0$. Its centre is $(1, 3)$ and radius $r_{1} = \sqrt{1^{2}+3^{2}-\alpha} = \sqrt{10-\alpha}$.The i

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$12$

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(C) The circle $c_{1}$ is $x^{2}+y^{2}-2x-6y+\alpha=0$. Its centre is $(1, 3)$ and radius $r_{1} = \sqrt{1^{2}+3^{2}-\alpha} = \sqrt{10-\alpha}$.The image of the centre $(1, 3)$ in the line $x-y+1=0$ is $(x_{1}, y_{1})$ given by $\frac{x_{1}-1}{1} = \frac{y_{1}-3}{-1} = -2 \frac{1-3+1}{1^{2}+(-1)^{2}} = -2 \frac{-1}{2} = 1$.Thus,$x_{1}-1=1 \Rightarrow x_{1}=2$ and $y_{1}-3=-1 \Rightarrow y_{1}=2$. The centre of $c_{2}$ is $(2, 2)$.The circle $c_{2}$ is given as $5x^{2}+5y^{2}+10gx+10fy+38=0$,which is $x^{2}+y^{2}+2gx+2fy+\frac{38}{5}=0$.Comparing with the standard form,the centre is $(-g, -f) = (2, 2)$,so $g=-2, f=-2$.The radius $r$ of $c_{2}$ is $\sqrt{g^{2}+f^{2}-\frac{38}{5}} = \sqrt{4+4-\frac{38}{5}} = \sqrt{8-\frac{38}{5}} = \sqrt{\frac{40-38}{5}} = \sqrt{\frac{2}{5}}$.Since reflection preserves the radius,$r_{1}^{2} = r^{2} \Rightarrow 10-\alpha = \frac{2}{5}$.Thus,$\alpha = 10 - \frac{2}{5} = \frac{48}{5}$.Finally,$\alpha+6r^{2} = \frac{48}{5} + 6(\frac{2}{5}) = \frac{48+12}{5} = \frac{60}{5} = 12$.

Let the mirror image of a circle $c_{1}: x^{2}+y^{2}-2x-6y+\alpha=0$ in the line $y=x+1$ be $c_{2}: 5x^{2}+5y^{2}+10gx+10fy+38=0$. If $r$ is the radius of circle $c_{2}$,then $\alpha+6r^{2}$ is equal to:

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