Let
A = $\{(x, y) \in R \times R | 2x^2 + 2y^2 - 2x - 2y = 1\}$,
B = $\{(x, y) \in R \times R | 4x^2 + 4y^2 - 16y + 7 = 0\}$ and
C = $\{(x, y) \in R \times R | x^2 + y^2 - 4x - 2y + 5 \le r^2\}$.
Then the minimum value of $|r|$ such that $A \cup B \subseteq C$ is
Show Hint
To ensure one circle completely contains another, always use:
\[
\text{Required radius} = \text{distance between centers} + \text{radius of inner circle}.
\]
For multiple sets, take the maximum of all such values.
We first interpret each set geometrically. Step 1: Identify set A \[ 2x^2 + 2y^2 - 2x - 2y = 1 \;\Rightarrow\; x^2 + y^2 - x - y = \frac{1}{2}. \] Completing squares: \[ (x-\tfrac{1}{2})^2 + (y-\tfrac{1}{2})^2 = 1. \] Hence, A is a circle with \[ \text{Center } C_A = \left(\tfrac{1}{2}, \tfrac{1}{2}\right), \quad \text{Radius } r_A = 1. \] Step 2: Identify set B \[ 4x^2 + 4y^2 - 16y + 7 = 0 \;\Rightarrow\; x^2 + y^2 - 4y = -\tfrac{7}{4}. \] Completing squares: \[ x^2 + (y-2)^2 = \tfrac{9}{4}. \] Hence, B is a circle with \[ \text{Center } C_B = (0,2), \quad \text{Radius } r_B = \tfrac{3}{2}. \] Step 3: Identify set C \[ x^2 + y^2 - 4x - 2y + 5 \le r^2. \] Completing squares: \[ (x-2)^2 + (y-1)^2 \le r^2. \] So, C is a circular disk with \[ \text{Center } C_C = (2,1), \quad \text{Radius } |r|. \] Step 4: Containment condition The condition \( A \cup B \subseteq C \) means that disk C must completely contain both circles A and B. For a circle with center \(C_1\) and radius \(r_1\) to contain another circle with center \(C_2\) and radius \(r_2\), \[ r_1 \ge d(C_1,C_2) + r_2. \] Step 5: Radius required to contain A \[ d(C_C,C_A) = \sqrt{\left(2-\tfrac12\right)^2 + \left(1-\tfrac12\right)^2} = \sqrt{\tfrac{10}{4}} = \tfrac{\sqrt{10}}{2}. \] Required radius: \[ |r| \ge \tfrac{\sqrt{10}}{2} + 1 = \tfrac{\sqrt{10}+2}{2}. \] Step 6: Radius required to contain B \[ d(C_C,C_B) = \sqrt{(2-0)^2 + (1-2)^2} = \sqrt{5}. \] Required radius: \[ |r| \ge \sqrt{5} + \tfrac{3}{2} = \tfrac{2\sqrt{5}+3}{2}. \] Step 7: Minimum required radius To contain both A and B, \[ |r| = \max\!\left\{ \tfrac{\sqrt{10}+2}{2},\; \tfrac{2\sqrt{5}+3}{2} \right\}. \] Since \[ \tfrac{2\sqrt{5}+3}{2}>\tfrac{\sqrt{10}+2}{2}, \] the minimum value of \(|r|\) is \[ \boxed{\tfrac{3+2\sqrt{5}}{2}}. \]
