Let {A}=\{({x}, {y}) \\in{R} \\times{R} \|2 {x}^{2} +2 {y}^{2}-2 {x} -2 {y}=1\} {B}=\{({x}, {y}) \\in{R} \\times{R} \|4 {x}^{2}+4 {y}^{2}-16 {y} +7=0\} and {C}=\{({x}, {y}) \\in{R} \\times{R} \|{x}^{2}+{y}^{2}-4 {x}-2 {y} +5 \\leq {r}^{2}\} Then the minimum value of |{r}| such that {A} \\cup{B} \\subseteq{C} is equal to [27 Jul 2021 Shift 1]

Circles Part 1

Section: Mathematics

Question : 11 of 100

Marks: +1, -0

Let
A={(x,y)∈R×R|2x2+2y2−2x −2y=1}
B={(x,y)∈R×R|4x2+4y2−16y +7=0} and
C={(x,y)∈R×R|x2+y2−4x−2y +5≤r2}
Then the minimum value of |r| such that A∪B⊆C is equal to

[27 Jul 2021 Shift 1]

3+√10
2
2+√10
2
3+2√5
2
1+√5

Validate

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