If the circles x² + y² - 16x - 20y + 164 = r² and (x-4)² + (y - 7)² = 36 intersect at two distinct p

Q: Wait, why do we use the absolute value $|R_1 - R_2|$ in the inequality?

Great observation! We use the absolute value because we don't know which circle is larger. The condition $|R_1 - R_2| < d$ ensures that the circles are not contained within one another. If $d$ were smaller than the difference of the radii, one circle would be floating inside the other without touching the boundary. The absolute value keeps the math robust regardless of which radius is bigger.

Q: Why can't we just say $r > 1$ and be done with it?

That is a common trap! While the algebra gives us $r > -1$, we must remember the physical reality of the problem. $r$ represents the radius of a circle, which is a distance. A distance cannot be negative. Therefore, we must impose the physical constraint $r > 0$. When we intersect $r > 0$ with our algebraic result $r > -1$, the tighter constraint $r > 0$ remains, but we must also respect the upper bound $r < 11$ derived from the other side of the inequality.

Q: What happens if $d = R_1 + R_2$ or $d = |R_1 - R_2|$?

Those are the boundary conditions! If $d = R_1 + R_2$, the circles are externally tangent (touching at exactly one point). If $d = |R_1 - R_2|$, they are internally tangent (also touching at one point). Since the problem explicitly demands two distinct intersection points, we must use strict inequalities ($<$) to exclude these tangency cases.

Question

If the circles $x^2 + y^2 - 16x - 20y + 164 = r^2$ and $(x-4)^2 + (y - 7)^2 = 36$ intersect at two distinct points, then:

Options:

(A) $0 < r < 1$

(B) $1 < r < 11$

(C) $r > 11$

(D) $r = 11$

Sigma Percentile · Accepted Answer

### The Geometry of Intersection: A Journey into Coordinate Space Welcome, fellow explorers of mathematics! Today, we are not just solving an equation; we are visualizing the dance of two circles in a coordinate plane. Imagine two circular ripples on a pond. When they overlap, they create a beautiful, lens-shaped region. The problem asks us to find the range of a parameter $r$ that allows this overlap to happen at exactly two distinct points. This is a classic JEE Advanced problem because it forces us to bridge the gap between pure algebra and geometric intuition. ### Phase 1: The Geometric Master Key Before we touch a single variable, let's establish our ground rules. For two circles with radii $R_1$ and $R_2$ and a distance $d$ between their centers, the condition for them to intersect at two distinct points is governed by the triangle inequality. The distance $d$ must be strictly less than the sum of the radii ($R_1 + R_2$) to ensure they are close enough to touch, and strictly greater than the absolute difference of the radii ($|R_1 - R_2|$) to ensure they aren't swallowed by one another. Our master inequality is: $$|R_1 - R_2| < d < R_1 + R_2$$ This is the compass that will guide us through the entire problem. ### Phase 2: Decoding the Circles We are given the first circle in a somewhat disguised form: $x^2 + y^2 - 16x - 20y + 164 = r^2$. To understand its soul—its center and its radius—we must perform a 'geometric makeover' using the method of completing the square. We group the $x$ terms and the $y$ terms: $(x^2 - 16x) + (y^2 - 20y) + 164 = r^2$. To complete the square for $x$, we take half of $-16$, which is $-8$, and square it to get $64$. For $y$, we take half of $-20$, which is $-10$, and square it to get $100$. We add and subtract these values: $(x^2 - 16x + 64) - 64 + (y^2 - 20y + 100) - 100 + 164 = r^2$. Notice the elegance here: $-64 - 100 + 164$ equals zero! The constants vanish, leaving us with the standard form: $$(x-8)^2 + (y-10)^2 = r^2$$ Now, the circle reveals its secrets: the center $C_1$ is at $(8, 10)$ and the radius $R_1$ is $r$. The second circle is already in its standard form: $(x-4)^2 + (y-7)^2 = 36$. Its center $C_2$ is $(4, 7)$ and its radius $R_2$ is $\sqrt{36} = 6$. ### Phase 3: The Distance Bridge Now that we have our centers, $C_1(8, 10)$ and $C_2(4, 7)$, we calculate the distance $d$ between them using the distance formula: $$d = \sqrt{(8-4)^2 + (10-7)^2}$$ This simplifies to $d = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$. The distance between our two centers is exactly $5$ units. We have all the components: $R_1 = r$, $R_2 = 6$, and $d = 5$. ### Phase 4: The Inequality Dance We return to our master inequality: $|r - 6| < 5 < r + 6$. We must solve this in two parts. First, the right side: $5 < r + 6$, which implies $r > -1$. However, as we discussed, $r$ is a radius, so $r > 0$. Second, the left side: $|r - 6| < 5$. This absolute value inequality expands to $-5 < r - 6 < 5$. Adding $6$ to all parts, we get $1 < r < 11$. Combining these, we find that the valid range for $r$ is **$1 < r < 11$**. ### Conclusion Look at what we have achieved! By translating the algebraic equations into geometric conditions, we have navigated the constraints of the problem. The range **$1 < r < 11$** is not just a set of numbers; it is the precise 'zone of existence' where these two circles will always embrace at two distinct points. Keep this mindset—always visualize the geometry behind the algebra—and you will conquer any problem JEE throws your way.

Animated Solution for Mathematics - Circles: If the circles $x^{2} + y^{2} - 16 x - 20 y + 164 = r^{2}$ and $(x - 4)^{2} + (y - 7)^{2} = 36$ intersect at two distinct points, then:

Select Answer:

Visualized Solution

Problem Visualization

Intersection Condition

Standard Form of Circle 1

Completing the Square

Center and Radius of Circle 1

Center and Radius of Circle 2

Distance Between Centers

Calculating $d$

Applying the Condition

Solving $r + 6 > 5$

Solving $∣ r - 6∣ < 5$

Final Range of $r$

The Sigma Insight: Standard and General Equation of a Circle

The Geometry of Intersection

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

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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 x2+y2−16x−20y+164=r2 and (x−4)2+(y−7)2=36 intersect at two distinct points, then: