1f:I[80107,["/_next/static/chunks/235ad55241b6443f.js","/_next/static/chunks/8337ab1f037b6654.js","/_next/static/chunks/c9e933b5fd1a0afc.js","/_next/static/chunks/848a330faa2461ac.js","/_next/static/chunks/9f2942e368a262d2.js","/_next/static/chunks/92f9e11244990470.js","/_next/static/chunks/a8b867fbb1dfca1a.js","/_next/static/chunks/1100f0390a901fa7.js","/_next/static/chunks/bbe6bd9d3c7b0355.js","/_next/static/chunks/a82195c3e5de7dd4.js"],"default"] 20:I[76763,["/_next/static/chunks/235ad55241b6443f.js","/_next/static/chunks/8337ab1f037b6654.js","/_next/static/chunks/c9e933b5fd1a0afc.js","/_next/static/chunks/848a330faa2461ac.js","/_next/static/chunks/9f2942e368a262d2.js","/_next/static/chunks/92f9e11244990470.js","/_next/static/chunks/a8b867fbb1dfca1a.js","/_next/static/chunks/1100f0390a901fa7.js","/_next/static/chunks/bbe6bd9d3c7b0355.js","/_next/static/chunks/a82195c3e5de7dd4.js"],"default"] 1e:T43a,{"@context":"https://schema.org","@graph":[{"@type":"BreadcrumbList","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https://www.divinejee.com"},{"@type":"ListItem","position":4,"name":"Consider a circle C1 x2y24 x2 y alpha5 Let its mirror image ","item":"https://www.divinejee.com/question/consider-a-circle-c1-x2-y24-x2-y-alpha5-let-its-mirror-image-in-the-line-y2-x-1-be-another-circle"}]},{"@type":"QAPage","mainEntity":{"@type":"Question","name":"Question Consider a circle C1 x2y24 x2 y alpha5 Let its mirror image on Topic","text":"

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 ...","answerCount":1,"upvoteCount":0,"datePublished":"2026-01-16","author":{"@type":"Organization","name":"DivineJEE"},"acceptedAnswer":{"@type":"Answer","text":"Check the detailed solution on the page.","upvoteCount":0,"url":"https://www.divinejee.com/question/consider-a-circle-c1-x2-y24-x2-y-alpha5-let-its-mirror-image-in-the-line-y2-x-1-be-another-circle"}}}]}21:T4a9,We have,

$$ \begin{aligned} & C_1: x^2+y^2-4 x-2 y=\alpha-5 \\\\ & C_1:(x-2)^2+(y-1)^5-5=\alpha-5 \\\\ & C_1:(x-2)^2+(y-1)^2=(\sqrt{\alpha})^2 \end{aligned} $$

So, centre and radius of $C_1$ are $(2,1)$ and $\sqrt{\alpha}$ respectively

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

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

$$ \begin{aligned} & \Rightarrow \frac{x-2}{2}=\frac{y-1}{-1}=\frac{-8}{5} \\\\ & \Rightarrow x=\frac{-6}{5} \text { and } y=\frac{13}{5} \end{aligned} $$

Now, $\left(\frac{-6}{5}, \frac{13}{5}\right)$ will be the centre of $C_2$

$$ \therefore f=\frac{6}{5} \text { and } g=\frac{-13}{5} $$

Now, radius of $\mathrm{C}_2=r=\sqrt{f^2+g^2-\frac{36}{5}}$

$$ \begin{aligned} & \Rightarrow r=\sqrt{\frac{36}{25}+\frac{169}{25}-\frac{36}{5}}=1 \\\\ & \because r=1 \text { so, } \alpha=1 \\\\ & \therefore \alpha+r=1+1=2 \end{aligned} $$

Concept :

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

$$ \frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{-2\left(a x_1+b y_1+c\right)}{\left(a^2+b^2\right)} $$