Question

Let $A=\left\{\right(x,y)\in R\times R|2x^2+2y^2-2x-2y=1\},$
$B=\left\{\right(x,y)\in R\times R|4x^2+4y^2-16y+7=0\}$ and
$C=\left\{\right(x,y)\in R\times R|x^2+y^2-4x-2y+5\le r^2\}$.
Then the minimum value of $|r|$ such that $A\cup B\subseteq C$ is equal to:

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

Answer 1

The correct option is B $\frac{3+2\sqrt{5}}{2}$

$A=\left\{\right(x,y)\in R\times R|2x^2+2y^2-2x-2y=1\},$
Equation of circle is $x^2+y^2-x-y-\frac{1}{2}=0$
Centre of circle, $C_1\equiv (\frac{1}{2},\frac{1}{2})$
and radius, $r_1=1$

$B=\left\{\right(x,y)\in R\times R|4x^2+4y^2-16y+7=0\}$
Equation of circle is $x^2+y^2-4y+\frac{7}{4}=0$
Centre of circle, $C_2\equiv (0,2)$
and radius, $r_2=\frac{3}{2}$

$C=\left\{\right(x,y)\in R\times R|x^2+y^2-4x-2y+5\le r^2\}$
Equation of circle is $(x-2{)}^2+(y-1{)}^2\le r^2$
Centre of circle, $C_3\equiv (2,1)$
and radius, $r_3=|r|$



From diagram it is clear that $C_3{C}_2\le |r-\frac{3}{2}|$
$\Rightarrow |r-\frac{3}{2}|\ge \sqrt{5}$

Case $\left(1\right):$
$r-\frac{3}{2}\ge \sqrt{5}$
$\Rightarrow r\ge \frac{3+2\sqrt{5}}{2}$

Case $\left(2\right):$
$r-\frac{3}{2}\le -\sqrt{5}$
$\Rightarrow r\le \frac{3-2\sqrt{5}}{2}$
Which is not possible.

$\therefore r_{min}=\frac{3+2\sqrt{5}}{2}$