Let $A=\left\{(x, y) \in R \times R \mid 2 x^{2}+2 y^{2}-2 x-2 y=1\right\}$, $B=\left\{(x, y) \in R \times R \mid 4 x^{2}+4 y^{2}-16 y+7=0\right\}$ and $C=\left\{(x, y) \in R \times R \mid x^{2}+y^{2}-4 x-2 y+5 \leq r^{2}\right\}$. Then the minimum value of $|r|$ such that $A \cup B \subseteq C$ is equal to :

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Circle

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 :

Answer: (B)

A

is the circle

x^{2} + y^{2} - x - y = \frac{1}{2}

, with centre

P_{1} (\frac{1}{2}, \frac{1}{2})

and radius

1

B

is the circle

x^{2} + y^{2} - 4 y + \frac{7}{4} = 0

, with centre

P_{2} (0, 2)

and radius

\frac{3}{2}

C

is the disc with centre

P_{3} (2, 1)

and radius

∣ r ∣

.
A circle of centre

P

and radius

ρ

lies inside that disc exactly when

P P_{3} + ρ \leq ∣ r ∣

.
For

A

P_{1} P_{3} = (\frac{3}{2})^{2} + (\frac{1}{2})^{2} = \frac{10}{2}

, so

∣ r ∣ \geq 1 + \frac{10}{2} = \frac{2 + 10}{2} \approx 2.58

.
For

B

P_{2} P_{3} = 4 + 1 = 5

, so

∣ r ∣ \geq \frac{3}{2} + 5 = \frac{3 + 2 5}{2} \approx 3.74

.
Both conditions must hold, so the least value of

∣ r ∣

\frac{3 + 2 5}{2}

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Single correct option