Question

Let A {(x, y) ∈ x R | 2x² 2y² - 2x - 2y  =1},  

B {(x, y) ∈ R x R | 4x² 4y² - 16y + 7 = 0} and 

C = {(x, y) ∈ R x R | X² + y² - 4x - 2y + 5 ≤ r²

Then the minimum value of r such that A \(\cup\) B \(\subseteq\) C is equal to

(1) \(\frac{3 + \sqrt{10}}{2}\)

(2) \(\frac{2+\sqrt{10}}{2}\)

(3) \(\frac{3 + 2\sqrt{5}}{2}\)

(4) 1 + \(\sqrt{5}\)

Answer 1

Correct option (3) \(\frac{3 + 2\sqrt{5}}{2}\)

S₁ : x² + y² - x - y - \(\frac{1}{2}\) = 0;  C₁ \(\left(\frac{1}{2}, \frac{1}{2} \right)\)

r₁ = \(\sqrt{\frac{1}{4} + \frac{1}{4} + \frac{1} {2}} = 1\)

S₂ : x² + y² - 4y + \(\frac{7}{4}\) = 0; C_{2 :} (0,2)

r₂ = \(\sqrt{4 - \frac{7}{4}} = \frac{3}{2}\)

S₃ = x² + y² - 4x - 2y + 5 - r² = 0

C₃ : (2,1)

r₃ = \(\sqrt{4 + 1 - 5 + r^2}\) = |r|