If the circles (x + 1)² + (y + 2)² = r² and x² + y² - 4x - 4y + 4 = 0 intersect at exactly two disti

Q: Wait, why do we use the inequality $|r_1 - r_2| < d < r_1 + r_2$?

This inequality is the geometric soul of circle intersection! If $d > r_1 + r_2$, the circles are too far apart and don't touch. If $d < |r_1 - r_2|$, one circle is completely inside the other without touching. The range $|r_1 - r_2| < d < r_1 + r_2$ is the 'sweet spot' where the circles must overlap to create exactly two intersection points.

Q: Why do we reject the negative values for $r$?

Great catch! In the inequality $|r - 2| 0$. When we combine this with our other condition $r > 3$, the lower bound becomes $3$.

Q: What happens exactly at $r = 3$ and $r = 7$?

At $r = 3$, the circles touch externally at a single point. At $r = 7$, the circles touch internally at a single point. Since the problem asks for *exactly two distinct points*, we must exclude these boundary values, which is why we use strict inequalities ($3 < r < 7$).

Question

If the circles $(x + 1)^2 + (y + 2)^2 = r^2$ and $x^2 + y^2 - 4x - 4y + 4 = 0$ intersect at exactly two distinct points, then

Options:

(A) $5 < r < 9$

(B) $0 < r < 7$

(C) $3 < r < 7$

(D) $\frac{1}{2} < r < 7$

Sigma Percentile · Accepted Answer

### The Dance of Two Circles: A Geometric Odyssey Imagine you are standing on a vast, flat coordinate plane. You have two circles, and you are tasked with a delicate mission: finding the exact range of the radius $r$ of the first circle so that it dances with the second circle, intersecting it at exactly two distinct points. This is not just algebra; it is a story of proximity, boundaries, and the beautiful logic of geometry. ### Phase 1: Unmasking the Circles First, let us look at our protagonists. The first circle is given by the equation $(x + 1)^2 + (y + 2)^2 = r^2$. This is already in the standard form $(x - h)^2 + (y - k)^2 = R^2$, which is a gift! We can immediately see that its center, $C_1$, is anchored at $(-1, -2)$, and its radius is $r$. Now, consider the second circle: $x^2 + y^2 - 4x - 4y + 4 = 0$. It is a bit shy, hiding in its general form. To understand it, we must complete the square. By grouping the $x$ and $y$ terms, we get $(x^2 - 4x + 4) + (y^2 - 4y + 4) = 4$. This simplifies beautifully to: $$ (x - 2)^2 + (y - 2)^2 = 2^2 $$ Now, the truth is revealed: its center $C_2$ is at $(2, 2)$, and it has a fixed radius of $2$. ### Phase 2: The Bridge Between Centers To understand how these two circles interact, we must measure the distance between their hearts—their centers. Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$, we calculate the distance between $C_1(-1, -2)$ and $C_2(2, 2)$: $$ d = \sqrt{(2 - (-1))^2 + (2 - (-2))^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5 $$ So, the centers are exactly $5$ units apart. This distance $d = 5$ is the bridge that dictates their relationship. ### Phase 3: The Condition for Intersection Here is the core of our journey. For two circles to intersect at exactly two distinct points, they must be close enough to touch but not so close that one is swallowed by the other. The condition is defined by the inequality: $$ |r_1 - r_2| < d < r_1 + r_2 $$ Substituting our values $r_1 = r$, $r_2 = 2$, and $d = 5$, we get: $$ |r - 2| < 5 < r + 2 $$ ### Phase 4: Solving the Inequality This compound inequality is our final hurdle. Let us break it into two manageable parts. First, consider the right side: $5 < r + 2$. Subtracting $2$ from both sides, we find $r > 3$. This makes perfect sense! If $r$ were $3$, the circles would touch externally. To have two intersection points, the radius must be larger than $3$. Second, consider the left side: $|r - 2| < 5$. This implies that $r - 2$ must be between $-5$ and $5$: $$ -5 < r - 2 < 5 $$ Adding $2$ to all parts, we get $-3 < r < 7$. This upper bound of $7$ is fascinating—if $r$ were $7$, the first circle would be so large that it would touch the second circle internally. We must stay below this limit. ### The Grand Finale Finally, we combine our conditions: $r > 3$ and $-3 < r < 7$. Since $r$ must be a positive radius, we ignore the negative part of the second interval. The intersection of these two conditions is the elegant range: $$ 3 < r < 7 $$ And there you have it! By visualizing the geometry and carefully applying the conditions of intersection, we have found the exact range where these two circles perform their perfect, two-point dance. **Geometry is not just about shapes; it is about understanding the relationships between them.**

Animated Solution for Mathematics - Circles: If the circles $(x + 1)^{2} + (y + 2)^{2} = r^{2}$ and $x^{2} + y^{2} - 4 x - 4 y + 4 = 0$ intersect at exactly two distinct points, then

Select Answer:

Visualized Solution

Visualize the Geometry

Analyze Circle 1

Analyze Circle 2

Distance Between Centers

Condition for Intersection

Solving the First Inequality

Solving the Second Inequality

Final Range of $r$

The Sigma Insight: Standard and General Equation of a Circle

The Dance of Two Circles

Phase 1

Phase 2

Phase 3

Phase 4

The Grand Finale

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Animated Solution for Mathematics - Circles: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the circles (x+1)2+(y+2)2=r2 and x2+y2−4x−4y+4=0 intersect at exactly two distinct points, then