Let A={(x, y) in R times R mid 2 x^{2}+2 y^{2 | vedclass

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(D) The set $A$ represents a circle $S_1: x^2 + y^2 - x - y - \frac{1}{2} = 0$. Its center $C_1 = (\frac{1}{2}, \frac{1}{2})$ and radius $r_1 = \sqrt{

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$\frac{3+2 \sqrt{5}}{2}$

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(D) The set $A$ represents a circle $S_1: x^2 + y^2 - x - y - \frac{1}{2} = 0$. Its center $C_1 = (\frac{1}{2}, \frac{1}{2})$ and radius $r_1 = \sqrt{(\frac{1}{2})^2 + (\frac{1}{2})^2 + \frac{1}{2}} = \sqrt{\frac{1}{4} + \frac{1}{4} + \frac{1}{2}} = 1$.The set $B$ represents a circle $S_2: x^2 + y^2 - 4y + \frac{7}{4} = 0$. Its center $C_2 = (0, 2)$ and radius $r_2 = \sqrt{0^2 + 2^2 - \frac{7}{4}} = \sqrt{4 - \frac{7}{4}} = \sqrt{\frac{9}{4}} = \frac{3}{2}$.The set $C$ represents a disk $S_3: (x-2)^2 + (y-1)^2 \leq r^2$. Its center $C_3 = (2, 1)$ and radius $R = |r|$.For $A \subseteq C$,the distance between centers $C_1 C_3 + r_1 \leq R$. $C_1 C_3 = \sqrt{(2 - \frac{1}{2})^2 + (1 - \frac{1}{2})^2} = \sqrt{(\frac{3}{2})^2 + (\frac{1}{2})^2} = \sqrt{\frac{9}{4} + \frac{1}{4}} = \sqrt{\frac{10}{4}} = \frac{\sqrt{10}}{2}$.So,$R \geq \frac{\sqrt{10}}{2} + 1 = \frac{2 + \sqrt{10}}{2}$.For $B \subseteq C$,the distance between centers $C_2 C_3 + r_2 \leq R$. $C_2 C_3 = \sqrt{(2 - 0)^2 + (1 - 2)^2} = \sqrt{2^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5}$.So,$R \geq \sqrt{5} + \frac{3}{2} = \frac{3 + 2\sqrt{5}}{2}$.Since $A \cup B \subseteq C$,$R$ must satisfy both conditions. Thus,$R \geq \max(\frac{2 + \sqrt{10}}{2}, \frac{3 + 2\sqrt{5}}{2})$.Comparing the two values,$\frac{3 + 2\sqrt{5}}{2} \approx \frac{3 + 4.47}{2} = 3.735$ and $\frac{2 + \sqrt{10}}{2} \approx \frac{2 + 3.16}{2} = 2.58$.Therefore,the minimum value is $\frac{3 + 2\sqrt{5}}{2}$.

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

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