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QJEEM23Ma95c0: JEE Main 2023 Mathematics - Class 11 JEEM23Ma95c0 Solution: Consider a circle . Let its mirror image in the line be another …

Consider a circle C1:x2+y2−4x−2y=α−5C_{1}: x^{2}+y^{2}-4 x-2 y=\alpha-5 . Let its mirror image in the line y=2x+1y=2 x+1 be another circle C2:5x2+5y2−10fx−10gy+36=0C_{2}: 5 x^{2}+5 y^{2}-10 f x-10 g y+36=0 . Let rr be the radius of C2C_{2} . Then α+r\alpha+r is equal to _________.

Step-by-Step Verified Solution

We have,

amp;C1:x2+y2−4x−2y=α−5amp;C1:(x−2)2+(y−1)5−5=α−5amp;C1:(x−2)2+(y−1)2=(α)2\begin{aligned}& C_1: x^2+y^2-4 x-2 y=\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{aligned}

So, centre and radius of C1C_1 are (2,1)(2,1) and α\sqrt{\alpha} respectively

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

x−22=y−1−1=−2(4−1+1)22+(−1)2\frac{x-2}{2}=\frac{y-1}{-1}=\frac{-2(4-1+1)}{2^2+(-1)^2}

amp;⇒x−22=y−1−1=−85amp;⇒x=−65 and y=135\begin{aligned}& \Rightarrow \frac{x-2}{2}=\frac{y-1}{-1}=\frac{-8}{5} \\\\& \Rightarrow x=\frac{-6}{5} \text { and } y=\frac{13}{5}\end{aligned}

Now, (−65,135)\left(\frac{-6}{5}, \frac{13}{5}\right) will be the centre of C2C_2

∴f=65 and g=−135\therefore f=\frac{6}{5} \text { and } g=\frac{-13}{5}

Now, radius of C2=r=f2+g2−365\mathrm{C}_2=r=\sqrt{f^2+g^2-\frac{36}{5}}

amp;⇒r=3625+16925−365=1amp;∵r=1 so, α=1amp;∴α+r=1+1=2\begin{aligned}& \Rightarrow r=\sqrt{\frac{36}{25}+\frac{169}{25}-\frac{36}{5}}=1 \\\\& \because r=1 \text { so, } \alpha=1 \\\\& \therefore \alpha+r=1+1=2\end{aligned}

Concept :

Image of a point (x1,y1)\left(x_1, y_1\right) w.r.t. ax+by+c=0a x+b y+c=0 is (x,y)(x, y) , then

x−x1a=y−y1b=−2(ax1+by1+c)(a2+b2)\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{-2\left(a x_1+b y_1+c\right)}{\left(a^2+b^2\right)}

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