QJEEM21Mae1b8: JEE Main 2021 Mathematics - Class 11 JEEM21Mae1b8 Solution: Let , and . Then the minimum value of

Question

Verified Solution • 2025–26 Session moderate

Let

A = \{ (x,y) \in R \times R|2{x^2} + 2{y^2} - 2x - 2y = 1\}

,

B = \{ (x,y) \in R \times R|4{x^2} + 4{y^2} - 16y + 7 = 0\}

and

C = \{ (x,y) \in R \times R|{x^2} + {y^2} - 4x - 2y + 5 \le {r^2}\}

.

Then the minimum value of |r| such that

A \cup B \subseteq C

is equal to

Accepted Answer

{S_1}:{x^2} + {y^2} - x - y - {1 \over 2} = 0;{C_1}\left( {{1 \over 2},{1 \over 2}} \right)

{r_1} = \sqrt {{1 \over 4} + {1 \over 4} + {1 \over 2}} = 1

{S_2}:{x^2} + {y^2} - 4y + {7 \over 4} = 0;{C_2}:(0,2)

{r_2} = \sqrt {4 - {7 \over 4}} = {3 \over 2}

{S_3} = {x^2} + {y^2} - 4x - 2y + 5 - {r^2} = 0

C₃ (2, 1)

{r_3} = \sqrt {4 + 1 - 5 + {r^2}} = |r|

{C_1}{C_3} = \sqrt {{5 \over 2}}

\sqrt{\frac{5}{2}} \leq | r - 1 | \Rightarrow \begin{matrix} r \leq 1 + \sqrt{\frac{5}{2}} \\ r \geq \frac{3}{2} + \sqrt{5} \end{matrix}}

{C_2}{C_3} = \sqrt 5 \le \left| {r - {3 \over 2}} \right|

\begin{matrix} r - \frac{3}{2} \geq \sqrt{5} \\ r - \frac{3}{2} \leq - \sqrt{5} \end{matrix}}

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