Let A = \ (x, y) ∈ R × R ∣ 2 x ^ 2 + 2 y ^ 2 - 2x - 2y = 1\ B = \...

Given,

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

So, the centre is ${C}_{1}: \left(\dfrac{1}{2}, \dfrac{1}{2}\right) and the radius is${r}_{1} = \sqrt{\dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{2}} = 1$</p><p>${S}_{2}: {x}^{2} + {y}^{2} - 4y + \dfrac{7}{4} = 0$</p><p>So, the centre is${C}_{2}: \left(0, 2\right) and the radius is

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

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

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

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

From the diagram, we can say that if $A \cup B ⊆ C,$ then the circle $A \text{ and } B$ should lie in the circle $C$

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

So, \left\sqrt{\dfrac{5}{2}} \le \left|r - 1\right| ⇒ $$\begin{array}{ll}r \le 1 - \sqrt{\dfrac{5}{2}} \\ r \ge 1 + \sqrt{\dfrac{5}{2}}\end{array}$$\right\}$</p><p>Also,${C}_{2}{C}_{3} = \sqrt{5} \le \left|r - \dfrac{3}{2}\right|$</p><p>$⇒ \left$$\begin{array}{ll}r \ge \sqrt{5} + \dfrac{3}{2} \\ r \le \dfrac{3}{2} - \sqrt{5}\end{array}$$\right\}$</p><p>So, the minimum value of$r = \dfrac{3 + 2\sqrt{5}}{2}$