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Consider a circle $C_1: x^2+y^2-4 x-2 y=\alpha-5$.Let its mirror image in the line $y=2 x+1$ be another circle $C _2: 5 x ^2+5 y ^2-10 fx -10 gy +36=0$.Let $r$ be the radius of $C _2$. Then $\alpha+ r$ is equal to $......$.
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$4$
Solution
$x^2+y^2-4 x-2 y+5-\alpha=0,$
$C_1(2,1) r_1=\sqrt{\alpha}$
$2 x-y+1=0$
Image of $(2,1)$
$\frac{x-2}{2}=\frac{y-1}{-1}=\frac{-2(4-1+1)}{5}$
$\frac{x-2}{2}=\frac{y-1}{-1}=\frac{-8}{5}$
$x=2-\frac{16}{5}=\frac{-6}{5}, y=1+\frac{8}{5}=\frac{13}{5}$
$x^2+y^2-2 f x-2 g y+\frac{36}{5}=0$
$C_2(f, g)$
$r _2=\sqrt{ f ^2+ g ^2-\frac{36}{5}}$
$\alpha= f ^2+ g ^2-\frac{36}{5}$
$\therefore f =-\frac{6}{5}, g =\frac{13}{5}$
$\alpha=\frac{36}{25}+\frac{169}{25}-\frac{36}{5}$
$=\frac{36+169-180}{25} \Rightarrow \alpha=1 \Rightarrow r =1$
$\therefore \alpha+ r =2$
