JEE Main 2023 — Mathematics Coordinate Geometry

Given,

$${C}_{1}:{x}_{2}+{y}_{2}–4x–2y+(5–\alpha )=0$$C1​:x2​+y2​–4x–2y+(5–α)=0

So, its centre will be, ${O}_{1}=(2,1)$O1​=(2,1) and radius $=\sqrt{\alpha }$=α​

And ${C}_{2}:5{x}^{2}+5{y}^{2}–10fx–10gy+36=0$C2​:5x2+5y2–10fx–10gy+36=0

$$\Rightarrow {C}_{2}:{x}^{2}+{y}^{2}–2fx–2gy+\frac{36}{5}=0$$⇒C2​:x2+y2–2fx–2gy+536​=0

So, Centre ${O}_{2}=(f,g)$O2​=(f,g) and radius $r=\sqrt{{f}^{2}+{g}^{2}-\frac{36}{5}}$r=f2+g2−536​​

Also given ${O}_{2}$O2​ is reflection of ${O}_{1}$O1​ in $2x–y+1=0$2x–y+1=0, so image formula we get,

$$\Rightarrow \frac{f-2}{2}=\frac{g-1}{-1}=-2\cdot (\frac{2\times 2-1+1}{{2}^{2}+{1}^{2}})$$⇒2f−2​=−1g−1​=−2⋅(22+122×2−1+1​)

$\Rightarrow f=\frac{-6}{5}$⇒f=5−6​ and $g=\frac{13}{5}$g=513​

So, radius $r=\sqrt{{(\frac{-6}{5})}^{2}+{(\frac{13}{5})}^{2}-\frac{36}{5}}=\frac{\sqrt{25}}{5}=1$r=(5−6​)2+(513​)2−536​​=525​​=1

$\Rightarrow r=1$⇒r=1and $\alpha =1$α=1 as they both are same radius circle,

Hence, $r+\alpha =2$r+α=2.