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

Question

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 ${C}_{2}:5{x}^{2}+5{y}^{2}-10fx-10gy+36=0$. Let $r$ be the radius of ${C}_{2}$. Then $\alpha +r$ is equal to \_\_\_\_\_\_\_\_\_.

Answer: 2

Accepted Answer

We have,

$\begin{gathered} {C}_{1}:{x}^{2}+{y}^{2}-4x-2y=\alpha -5 \ {C}_{1}:(x-2{)}^{2}+(y-1{)}^{5}-5=\alpha -5 \ {C}_{1}:(x-2{)}^{2}+(y-1{)}^{2}=(\sqrt{\alpha }{)}^{2} \end{gathered}$

So, centre and radius of ${C}_{1}$ are $(2,1)$ and $\sqrt{\alpha }$ respectively

Now, image of $(2,1)$ along the line $y=2x+1$ is,

$\frac{x-2}{2}=\frac{y-1}{-1}=\frac{-2(4-1+1)}{{2}^{2}+(-1{)}^{2}}$

$\begin{gathered} \Rightarrow \frac{x-2}{2}=\frac{y-1}{-1}=\frac{-8}{5} \ \Rightarrow x=\frac{-6}{5}y=\frac{13}{5} \end{gathered}$

Now, $(\frac{-6}{5},\frac{13}{5})$ will be the centre of ${C}_{2}$

$	herefore f=\frac{6}{5}g=\frac{-13}{5}$

Now, radius of ${C}_{2}=r=\sqrt{{f}^{2}+{g}^{2}-\frac{36}{5}}$

$\begin{gathered} \Rightarrow r=\sqrt{\frac{36}{25}+\frac{169}{25}-\frac{36}{5}}=1 \ \because r=1\alpha =1 \ 	herefore \alpha +r=1+1=2 \end{gathered}$

Concept :

Image of a point $({x}_{1},{y}_{1})$ w.r.t. $ax+by+c=0$ is $(x,y)$, then

$\frac{x-{x}_{1}}{a}=\frac{y-{y}_{1}}{b}=\frac{-2(a{x}_{1}+b{y}_{1}+c)}{({a}^{2}+{b}^{2})}$

Solutions · 1