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Let
$A=\left\{(x, y) \in R \times R \mid 2 x^{2}+2 y^{2}-2 x-2 y=1\right\}$
$B=\left\{(x, y) \in R \times R \mid 4 x^{2}+4 y^{2}-16 y+7=0\right\} \text { and }$
$C=\left\{(x, y) \in R \times R \mid x^{2}+y^{2}-4 x-2 y+5 \leq r^{2}\right\}$
Then the minimum value of $|r|$ such that $A \cup B \subseteq C$ is equal to:
$\frac{3+\sqrt{10}}{2}$
$1+\sqrt{5}$
$\frac{2+\sqrt{10}}{2}$
$\frac{3+2 \sqrt{5}}{2}$
Solution

$\mathrm{S}_{1}: \mathrm{x}^{2}+\mathrm{y}^{2}-\mathrm{x}-\mathrm{y}-\frac{1}{2}=0 ; \mathrm{C}_{1}\left(\frac{1}{2}, \frac{1}{2}\right)$
$\mathrm{r}_{1}=\sqrt{\frac{1}{4}+\frac{1}{4}+\frac{1}{2}}=1$
$\mathrm{~S}_{2}: \mathrm{x}^{2}+\mathrm{y}^{2}-4 \mathrm{y}+\frac{7}{4}=0 ; \mathrm{C}_{2}:(0,2)$
$r_{2}=\sqrt{4-\frac{7}{4}}=\frac{3}{2}$
$\mathrm{~S}_{3}: \mathrm{x}^{2}+\mathrm{y}^{2}-4 \mathrm{x}-2 \mathrm{y}+5-\mathrm{r}^{2}=0$
$\mathrm{C}_{3}:(2,1)$
$r_{3}=\sqrt{4+1-5+\mathrm{r}^{2}}=|\mathrm{r}|$
$\mathrm{C}_{1} \mathrm{C}_{3}=\sqrt{\frac{5}{2}}$
$\sqrt{\frac{5}{2}} \leq|r-1| \Rightarrow r \leq 1+\sqrt{\frac{5}{2}}$
$\quad\quad\quad\quad\quad\quad\quad\quad r \geq \frac{3}{2}+\sqrt{5}$
$C_{2} C_{3}=\sqrt{5} \leq\left|r-\frac{3}{2}\right|$
$\sqrt{5} \leq\left|r-\frac{3}{2}\right| \Rightarrow r-\frac{3}{2} \geq \sqrt{5}$
$\quad\quad\quad\quad\quad\quad\quad\quad r-\frac{3}{2} \leq-\sqrt{5}$