MathematicsMedium127×since 2002Q3024Let A={(x,y)∈R×R∣2x2+2y2−2x−2y=1}A = \{ (x,y) \in R \times R|2{x^2} + 2{y^2} - 2x - 2y = 1\}A={(x,y)∈R×R∣2x2+2y2−2x−2y=1}, B={(x,y)∈R×R∣4x2+4y2−16y+7=0}B = \{ (x,y) \in R \times R|4{x^2} + 4{y^2} - 16y + 7 = 0\}B={(x,y)∈R×R∣4x2+4y2−16y+7=0} and C={(x,y)∈R×R∣x2+y2−4x−2y+5≤r2}C = \{ (x,y) \in R \times R|{x^2} + {y^2} - 4x - 2y + 5 \le {r^2}\}C={(x,y)∈R×R∣x2+y2−4x−2y+5≤r2}. Then the minimum value of |r| such that A∪B⊆CA \cup B \subseteq CA∪B⊆C is equal toA3+102{{3 + \sqrt {10} } \over 2}23+10B2+102{{2 + \sqrt {10} } \over 2}22+10C3+252{{3 + 2\sqrt 5 } \over 2}23+25D1+51 + \sqrt 51+5Check answerSkip