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

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Question

Image of centre $c _{1} \equiv(1,3)$ in $x - y +1=0$ is given by

$\frac{x_{1}-1}{1}=\frac{y_{1}-3}{-1}=\frac{-2(1-3+1)}{1^{2}+1^{2}}$

$x_{1}=2,

Vedclass · Accepted Answer

$12$

Vedclass · Answer

Image of centre $c _{1} \equiv(1,3)$ in $x - y +1=0$ is given by

$\frac{x_{1}-1}{1}=\frac{y_{1}-3}{-1}=\frac{-2(1-3+1)}{1^{2}+1^{2}}$

$x_{1}=2, y_{1}=2$

$	herefore$ Centre of circle $c _{2} \equiv(2,2)$

$	herefore$ Equation of $c_{2}$ be $x^{2}+y^{2}-4 x-4 y+\frac{38}{5}=0$

Now radius of $c _{2}$ is $\sqrt{4+4-\frac{38}{5}}=\sqrt{\frac{2}{5}}= r$

$\left(	ext { radius of } c _{1}ight)^{2}=\left(	ext { radius of } c _{2}ight)^{2}$

$10-\alpha=\frac{2}{5} \Rightarrow \alpha=\frac{48}{5}$

$	herefore \alpha+6 r ^{2}=\frac{48}{5}+\frac{12}{5}=12$

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

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