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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 f x-10 g y+36=0. Let r be the radius of C_2. Then \alpha+\mathrm{r} is equal to _____________.

Option: 1

2


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\begin{aligned} & x^2+y^2-4 x-2 y+5-\alpha=0 \\ & C_1(2,1) r_1=\sqrt{\alpha} \\ & 2 x-y+1=0 \\ & \text { 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) \end{aligned}

\begin{aligned} & \mathrm{r}_2=\sqrt{\mathrm{f}^2+\mathrm{g}^2-\frac{36}{5}} \\ & \alpha=\mathrm{f}^2+\mathrm{g}^2-\frac{36}{5} \\ & \therefore \mathrm{f}=-\frac{6}{5}, \mathrm{~g}=\frac{13}{5} \\ & \alpha=\frac{36}{25}+\frac{169}{25}-\frac{36}{5} \\ & =\frac{36+169-180}{25} \Rightarrow \alpha=1 \Rightarrow \mathrm{r}=1 \\ & \therefore \alpha+\mathrm{r}=2 \end{aligned}

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Pankaj Sanodiya

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