Let A = x , y ∈ R × R ∣ 2 x 2 + 2 y 2 - 2 x - 2 y = 1 B = x , y ∈ R × R ∣ 4 x 2 + 4 y 2 - 16 y + 7 = 0 and C = x , y ∈ R × R ∣ x 2 + y 2 - 4 x - 2 y + 5 ≤ r 2 . Then the minimum value of r such that A ∪ B ⊆ C is equal to

Step-by-step solution

Given, S 1 : x 2 + y 2 - x - y - 1 2 = 0 So, the centre is C 1 : 1 2 ,   1 2 and the radius is r 1 = 1 4 + 1 4 + 1 2 = 1 S 2 : x 2 + y 2 - 4 y + 7 4 = 0 So, the centre is C 2 : 0 , 2 and the radius is r 2 = 4 - 7 4 = 3 2 S 3 : x 2 + y 2 - 4 x - 2 y + 5 - r 2 = 0 So, the centre is C 3 : 2 , 1 and the radius is r 3 = 4 + 1 - 5 + r 2 = r From the diagram, we can say that if A ∪ B ⊆ C , then the circle A   &   B should lie in the circle C Here, C 1 C 3 = 5 2 So, 5 2 ≤ r - 1 &#8658