JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
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:
- A \(\frac{3+\sqrt{10}}{2}\)
- B \(1+\sqrt{5}\)
- C \(\frac{2+\sqrt{10}}{2}\)
- D \(\frac{3+2 \sqrt{5}}{2}\)
Answer & Solution
Correct Answer
(D) \(\frac{3+2 \sqrt{5}}{2}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{S}_{1}: \mathrm{x}^{2}+\mathrm{y}^{2}-\mathrm{x}-\mathrm{y}-\frac{1}{2}=0 ; \mathrm{C}_{1}\left(\frac{1}{2}, \frac{1}{2}\right)\) \(\mathrm{r}_{1}=\sqrt{\frac{1}{4}+\frac{1}{4}+\frac{1}{2}}=1\)…
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