Let A=(x,y)∈ R× Rmid 2x^2+2y^2-2x-2y=1… - JEE Main 2021 Mathematics PYQ | PrepForBharat

JEE Main 2021 — Mathematics Coordinate Geometry

2021mcqhard

Let

$A={(x,y)\in R\times R\mid 2{x}^{2}+2{y}^{2}-2x-2y=1}$

$B={(x,y)\in R\times R\mid 4{x}^{2}+4{y}^{2}-16y+7=0}$ and

$C={(x,y)\in R\times R\mid {x}^{2}+{y}^{2}-4x-2y+5\leq {r}^{2}}$ . Then the minimum value of $|r|$ such that $A\cup B\subseteq C$ is equal to

Official previous-year question

Held on 27 Jul 2021 · Verified 6 Jul 2026.

Options

A
$\frac{3+\sqrt{10}}{2}$
B
$\frac{2+\sqrt{10}}{2}$
C
$\frac{3+2\sqrt{5}}{2}$
D
$1+\sqrt{5}$

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Solution

Given,

${S}_{1}:{x}^{2}+{y}^{2}-x-y-\frac{1}{2}=0$

So, the centre is ${C}_{1}:(\frac{1}{2},\frac{1}{2})$ and the radius is ${r}_{1}=\sqrt{\frac{1}{4}+\frac{1}{4}+\frac{1}{2}}=1$

${S}_{2}:{x}^{2}+{y}^{2}-4y+\frac{7}{4}=0$

So, the centre is ${C}_{2}:(0,2)$ and the radius is

${r}_{2}=\sqrt{4-\frac{7}{4}}=\frac{3}{2}$

${S}_{3}:{x}^{2}+{y}^{2}-4x-2y+5-{r}^{2}=0$

So, the centre is ${C}_{3}:(2,1)$ and the radius is

${r}_{3}=\sqrt{4+1-5+{r}^{2}}=|r|$

JEE Main 2021 Mathematics, Circle — solution figure

From the diagram, we can say that if $A\cup B\subseteq C,$ then the circle $A&B$ should lie in the circle $C$

Here, ${C}_{1}{C}_{3}=\sqrt{\frac{5}{2}}$

So, $\sqrt{\frac{5}{2}}\leq |r-1|\Rightarrow \begin{matrix}r\leq 1-\sqrt{\frac{5}{2}} \\ r\geq 1+\sqrt{\frac{5}{2}}\end{matrix}}$

Also, ${C}_{2}{C}_{3}=\sqrt{5}\leq |r-\frac{3}{2}|$

$\Rightarrow \begin{matrix}r\geq \sqrt{5}+\frac{3}{2} \\ r\leq \frac{3}{2}-\sqrt{5}\end{matrix}}$

So, the minimum value of $r=\frac{3+2\sqrt{5}}{2}$

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