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

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Question

$\mathrm{S}_{1}: \mathrm{x}^{2}+\mathrm{y}^{2}-\mathrm{x}-\mathrm{y}-\frac{1}{2}=0 ; \mathrm{C}_{1}\left(\frac{1}{2}, \frac{1}{2}\right)$

$\mathrm{

Vedclass · Accepted Answer

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

Vedclass · Answer

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

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