JEE Main 2021 — Mathematics Coordinate Geometry

Given,

$${S}_{1}:{x}^{2}+{y}^{2}-x-y-\frac{1}{2}=0$$S1​:x2+y2−x−y−21​=0

So, the centre is ${C}_{1}:(\frac{1}{2},\frac{1}{2})$C1​:(21​,21​) and the radius is ${r}_{1}=\sqrt{\frac{1}{4}+\frac{1}{4}+\frac{1}{2}}=1$r1​=41​+41​+21​​=1

$${S}_{2}:{x}^{2}+{y}^{2}-4y+\frac{7}{4}=0$$S2​:x2+y2−4y+47​=0

So, the centre is ${C}_{2}:(0,2)$C2​:(0,2) and the radius is

$${r}_{2}=\sqrt{4-\frac{7}{4}}=\frac{3}{2}$$r2​=4−47​​=23​

$${S}_{3}:{x}^{2}+{y}^{2}-4x-2y+5-{r}^{2}=0$$S3​:x2+y2−4x−2y+5−r2=0

So, the centre is ${C}_{3}:(2,1)$C3​:(2,1) and the radius is

$${r}_{3}=\sqrt{4+1-5+{r}^{2}}=|r|$$r3​=4+1−5+r2​=∣r∣

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

Here, ${C}_{1}{C}_{3}=\sqrt{\frac{5}{2}}$C1​C3​=25​​

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, ${C}_{2}{C}_{3}=\sqrt{5}\leq |r-\frac{3}{2}|$C2​C3​=5​≤∣r−23​∣

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

So, the minimum value of $r=\frac{3+2\sqrt{5}}{2}$r=23+25​​