JEE MAIN - Mathematics (2021 - 27th July Morning Shift - No. 16)
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
Then the minimum value of |r| such that $$A \cup B \subseteq C$$ is equal to
$${{3 + \sqrt {10} } \over 2}$$
$${{2 + \sqrt {10} } \over 2}$$
$${{3 + 2\sqrt 5 } \over 2}$$
$$1 + \sqrt 5 $$
Explanation
$${S_1}:{x^2} + {y^2} - x - y - {1 \over 2} = 0;{C_1}\left( {{1 \over 2},{1 \over 2}} \right)$$
$${r_1} = \sqrt {{1 \over 4} + {1 \over 4} + {1 \over 2}} = 1$$
$${S_2}:{x^2} + {y^2} - 4y + {7 \over 4} = 0;{C_2}:(0,2)$$
$${r_2} = \sqrt {4 - {7 \over 4}} = {3 \over 2}$$
$${S_3} = {x^2} + {y^2} - 4x - 2y + 5 - {r^2} = 0$$
C3 (2, 1)
$${r_3} = \sqrt {4 + 1 - 5 + {r^2}} = |r|$$
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$${C_1}{C_3} = \sqrt {{5 \over 2}} $$
$$\sqrt {{5 \over 2}} \le |r - 1| \Rightarrow \left. \matrix{ r \le 1 + \sqrt {{5 \over 2}} \hfill \cr r \ge {3 \over 2} + \sqrt 5 \hfill \cr} \right\}$$
$${C_2}{C_3} = \sqrt 5 \le \left| {r - {3 \over 2}} \right|$$
$$\left. \matrix{ r - {3 \over 2} \ge \sqrt 5 \hfill \cr r - {3 \over 2} \le - \sqrt 5 \hfill \cr} \right\}$$
$${r_1} = \sqrt {{1 \over 4} + {1 \over 4} + {1 \over 2}} = 1$$
$${S_2}:{x^2} + {y^2} - 4y + {7 \over 4} = 0;{C_2}:(0,2)$$
$${r_2} = \sqrt {4 - {7 \over 4}} = {3 \over 2}$$
$${S_3} = {x^2} + {y^2} - 4x - 2y + 5 - {r^2} = 0$$
C3 (2, 1)
$${r_3} = \sqrt {4 + 1 - 5 + {r^2}} = |r|$$
$${C_1}{C_3} = \sqrt {{5 \over 2}} $$
$$\sqrt {{5 \over 2}} \le |r - 1| \Rightarrow \left. \matrix{ r \le 1 + \sqrt {{5 \over 2}} \hfill \cr r \ge {3 \over 2} + \sqrt 5 \hfill \cr} \right\}$$
$${C_2}{C_3} = \sqrt 5 \le \left| {r - {3 \over 2}} \right|$$
$$\left. \matrix{ r - {3 \over 2} \ge \sqrt 5 \hfill \cr r - {3 \over 2} \le - \sqrt 5 \hfill \cr} \right\}$$
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