Let A = {(x, y) ∈ R × R | 2x² + 2y² - 2x - 2y = 1}, B = {(x, y) ∈ R × R | 4x² + 4y² - 16y + 7 = 0} a

Q: Wait, why do we use the distance between centers plus the radius? Why not just the distance between centers?

Great question! If you only consider the distance between centers, you are only ensuring the *center* of the inner circle is inside the outer one. But a circle has 'width'. To ensure the *entire* circle is inside, the outer circle must extend far enough to cover the furthest point of the inner circle. That furthest point is exactly at a distance of $Dist(C_{inner}, C_{outer}) + r_{inner}$ from the outer center.

Q: Why do we take the maximum of the two values? Shouldn't it be the minimum?

Think of it as a 'tightest fit' problem. If we choose the smaller radius, we might successfully enclose circle $A$, but circle $B$ would be left partially outside. To satisfy the condition $A \cup B \subseteq C$, the container $C$ must be large enough to hold the 'hardest' circle to enclose. Therefore, we must pick the larger of the two required radii.

Q: Does the order of the union $A \cup B$ matter?

Not at all! The union $A \cup B$ is just the combined set of points from both circles. If $C$ contains $A$ and $C$ contains $B$, then by definition, $C$ contains their union. The logic holds regardless of which circle we analyze first.

Question

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

Options:

(A) $\frac{3+\sqrt{10}}{2}$

(B) $\frac{2+\sqrt{10}}{2}$

(C) $\frac{3+2\sqrt{5}}{2}$

(D) $1 + \sqrt{5}$

Sigma Percentile · Accepted Answer

### The Geometric Symphony

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving an equation; we are choreographing a dance of circles in the coordinate plane.

We have three sets, $A$, $B$, and $C$, defined by quadratic forms. Our mission is to find the minimum radius $|r|$ of a container, set $C$, that can perfectly encapsulate both $A$ and $B$.

Imagine you are an architect designing a circular enclosure that must safely house two smaller, pre-existing circular structures. This is the essence of our problem.

### Decoding the Circles

First, we must strip away the algebraic disguise of these sets. Let us look at set $A$: $2x^2 + 2y^2 - 2x - 2y = 1$.

To see the circle, we need the coefficients of $x^2$ and $y^2$ to be unity. Dividing by $2$, we get $x^2 + y^2 - x - y = \frac{1}{2}$.

By completing the square, we find the center $C_1 = (\frac{1}{2}, \frac{1}{2})$ and the radius $r_1 = 1$.

Now, look at set $B$: $4x^2 + 4y^2 - 16y + 7 = 0$. Normalizing this by dividing by $4$, we get $x^2 + y^2 - 4y + \frac{7}{4} = 0$.

This reveals a center $C_2 = (0, 2)$ and a radius $r_2 = \frac{3}{2}$.

Finally, the container $C$ is defined by $x^2 + y^2 - 4x - 2y + 5 \le r^2$, which simplifies to $(x-2)^2 + (y-1)^2 \le r^2$. This is a disk centered at $C_3 = (2, 1)$ with radius $|r|$.

### The Containment Logic

Here is the core of our journey. For a circle $A$ to be a subset of $C$, the distance between their centers, $Dist(C_1, C_3)$, plus the radius of $A$, must be less than or equal to the radius of $C$.

Mathematically, this is expressed as:

$$Dist(C_1, C_3) + r_1 \le |r|$$

This is the 'reach' of our container. If the container does not reach the furthest edge of the inner circle, it fails.

Let us calculate the distance:

$$Dist(C_1, C_3) = \sqrt{(2 - \frac{1}{2})^2 + (1 - \frac{1}{2})^2} = \sqrt{(\frac{3}{2})^2 + (\frac{1}{2})^2} = \sqrt{\frac{9}{4} + \frac{1}{4}} = \frac{\sqrt{10}}{2}$$

Thus, for $A$, we need:

$$|r| \ge \frac{\sqrt{10}}{2} + 1 = \frac{2+\sqrt{10}}{2}$$

### The Final Comparison

We repeat this for circle $B$. The distance is:

$$Dist(C_2, C_3) = \sqrt{(2 - 0)^2 + (1 - 2)^2} = \sqrt{4 + 1} = \sqrt{5}$$

The condition for $B$ is:

$$|r| \ge \sqrt{5} + \frac{3}{2} = \frac{3+2\sqrt{5}}{2}$$

Now, we stand at the crossroads. We have two constraints: $|r| \ge \frac{2+\sqrt{10}}{2}$ and $|r| \ge \frac{3+2\sqrt{5}}{2}$.

To satisfy both, we must choose the maximum of these two values. A quick mental check: $\sqrt{10} \approx 3.16$, so the first value is $\approx 2.58$.

Since $\sqrt{5} \approx 2.23$, the second value is $\approx 3.73$. The second constraint is clearly the stricter one.

Thus, the minimum radius required to capture both circles is **$\frac{3+2\sqrt{5}}{2}$**. You have successfully navigated the geometry, the algebra, and the logic. Well done!

Select Answer:

Visualized Solution

The Geometric Setup

Decoding Set $A$

Center and Radius of Circle $A$

Decoding Set $B$

Center and Radius of Circle $B$

Analyzing the Container Region $C$

The Subset Condition $A \cup B \subseteq C$

Enclosing Circle $A$

First Constraint on $∣ r ∣$

Enclosing Circle $B$

Second Constraint on $∣ r ∣$

Finding the Minimum $∣ r ∣$

Final Conclusion

The Sigma Insight: Standard and General Equation of a Circle

The Geometric Symphony

Decoding the Circles

The Containment Logic

The Final Comparison

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Select Answer:

Visualized Solution

The Geometric Setup

Decoding Set .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Center and Radius of Circle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Decoding Set .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Center and Radius of Circle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Analyzing the Container Region .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

The Subset Condition .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A∪B⊆C

Enclosing Circle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

First Constraint on .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣r∣

Enclosing Circle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Second Constraint on .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣r∣

Finding the Minimum .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣r∣