A circle C of radius 2 lies in the second quadrant and touches both the coordinate axes. Let r be th

Q: Wait, why is the center of the first circle $(-2, 2)$? Shouldn't it be $(2, 2)$?

Great question! Remember the coordinate plane. The second quadrant is where $x$ is negative and $y$ is positive. Since the circle touches the $y$-axis at a distance of $2$ from the origin in the negative $x$-direction, its $x$-coordinate must be $-2$. Similarly, touching the $x$-axis at a distance of $2$ in the positive $y$-direction gives it a $y$-coordinate of $2$. Hence, $(-2, 2)$ is correct.

Q: Why do we use the condition $|r_1 - r_2| < d < r_1 + r_2$?

This is the fundamental condition for two circles to intersect at exactly two points. The right side, $d |r_1 - r_2|$, ensures that one circle is not completely contained within the other without touching. If $d = r_1 + r_2$, they touch externally at one point; if $d = |r_1 - r_2|$, they touch internally at one point. We need the range strictly between these limits for two intersection points.

Q: When solving $|2 - r| -3$?

When you solve the inequality $|2 - r| r > -3$. While mathematically $r$ can be greater than $-3$, we are dealing with a radius, which must be a positive physical quantity ($r > 0$). Therefore, we only consider the part of the solution where $r$ is positive, leading to $0 < r < 7$.

Question

A circle C of radius 2 lies in the second quadrant and touches both the coordinate axes. Let r be the radius of a circle that has centre at the point (2, 5) and intersects the circle C at exactly two points. If the set of all possible values of r is the interval $(\alpha,\beta)$ then $3\beta-2\alpha$ is equal to :

Options:

(A) 10

(B) 15

(C) 12

(D) 14

Sigma Percentile · Accepted Answer

### Analyzing the Setup Imagine you are standing on a coordinate plane, looking at a circle nestled in the second quadrant. This is our Circle $C_1$. It is a perfect, stable entity with a radius of $r_1 = 2$. Because it is tangent to both the $x$ and $y$ axes in the second quadrant, its center $C_1$ is locked at $(-2, 2)$. Now, introduce a second, dynamic circle. Its center $C_2$ is fixed at $(2, 5)$, but its radius $r$ is a variable. Our goal is to find the range of $r$ that forces these two circles to intersect at exactly two points. ### The Bridge: Calculating the Distance Before we can analyze the intersection, we must understand the space between them. The distance $d$ between the centers $C_1(-2, 2)$ and $C_2(2, 5)$ is the bridge that connects our two circles. Using the distance formula, $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$, we calculate: $$d = \sqrt{(2 - (-2))^2 + (5 - 2)^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = 5$$ This distance of $5$ units serves as the static backdrop for our problem. ### The Core Condition: The Dance of Two Points For two circles to intersect at exactly two points, they must be neither too far apart nor one inside the other. The mathematical condition for this is: $$|r_1 - r_2| < d < r_1 + r_2$$ This inequality is the heartbeat of the problem. It defines the "sweet spot" where the circles overlap. Substituting our known values, $r_1 = 2$ and $d = 5$, we get the compound inequality: $$|2 - r| < 5 < 2 + r$$ ### Solving the Inequality: The Final Stretch We break this into two manageable pieces. First, consider the right side: $5 < 2 + r$. Subtracting $2$ from both sides, we find $r > 3$. This is our lower bound; if the radius is $3$ or less, the circles are too far apart to intersect at two points. Next, consider the left side: $|2 - r| < 5$. This modulus inequality expands to: $$-5 < 2 - r < 5$$ Subtracting $2$ from all parts, we get $-7 < -r < 3$. Multiplying by $-1$ and reversing the inequality signs, we obtain $7 > r > -3$. Since a radius must be positive, we focus on $r < 7$. Combining these results, we find the valid interval for the radius: $$3 < r < 7$$ ### Final Calculation From the interval $3 < r < 7$, we identify the bounds as $\alpha = 3$ and $\beta = 7$. The final calculation, $3\beta - 2\alpha$, becomes: $$3(7) - 2(3) = 21 - 6 = 15$$ We have successfully navigated the geometry and the algebra to reach our final answer of **15**.

Select Answer:

Visualized Solution

Identifying Circle $C$

The Second Circle's Properties

Distance Between Centers

Calculating the Distance

Condition for Two-Point Intersection

Substituting Values

Solving the Upper Bound

Solving the Lower Bound (Part 1)

Solving the Lower Bound (Part 2)

Finding the Interval $(α, β)$

Final Calculation

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Bridge

The Core Condition

Solving the Inequality

Final Calculation

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Select Answer:

Visualized Solution

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

The Second Circle's Properties

Distance Between Centers

Calculating the Distance

Condition for Two-Point Intersection

Substituting Values

Solving the Upper Bound

Solving the Lower Bound (Part 1)

Solving the Lower Bound (Part 2)

Finding the Interval .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (α,β)

Final Calculation

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Bridge

The Core Condition

Solving the Inequality

Final Calculation

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