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:

त्रिज्या 2 वाला एक वृत्त C दूसरे चतुर्थांश में स्थित है और दोनों निर्देशांक अक्षों को स्पर्श करता है। मान लीजिए कि, एक वृत्त की त्रिज्या है जिसका केंद्र बिंदु (2, 5) पर है और यह वृत्त C को ठीक दो बिंदुओं पर प्रतिच्छेद करता है। यदि r के सभी संभावित मानों का समुच्चय अंतराल ($\alpha$, $\beta$) है, तो $3\beta - 2\alpha$ बराबर है:

• A. 15
• B. 14
• C. 12
• D. 10

Answer 1

Answer: (A)

15

Answer 2

To solve this problem, we need to analyze the geometry of the two circles and the conditions for their intersection at exactly two points.

1. Determine the properties of Circle C:

• Radius of circle C, $R_C = 2$.
• It lies in the second quadrant.
• It touches both the coordinate axes.

For a circle of radius $R$ to touch both the coordinate axes in the second quadrant, its center must be at $(-R, R)$. Therefore, the center of circle C, let's call it $C_C$, is $(-2, 2)$. The equation of circle C is $(x - (-2))^2 + (y - 2)^2 = 2^2$, which simplifies to $(x+2)^2 + (y-2)^2 = 4$.

2. Determine the properties of the second circle:

• Its center, let's call it $C_S$, is $(2, 5)$.
• Its radius is $r$.

3. Calculate the distance between the centers of the two circles: Let $d$ be the distance between $C_C(-2, 2)$ and $C_S(2, 5)$. Using the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ $d = \sqrt{(2 - (-2))^2 + (5 - 2)^2}$ $d = \sqrt{(2+2)^2 + (3)^2}$ $d = \sqrt{4^2 + 3^2}$ $d = \sqrt{16 + 9}$ $d = \sqrt{25}$ $d = 5$.

4. Apply the condition for two circles to intersect at exactly two points: For two circles with radii $R_1$ and $R_2$ and distance between centers $d$ to intersect at exactly two points, the following condition must be satisfied: $|R_1 - R_2| < d < R_1 + R_2$

In our case, $R_1 = R_C = 2$, $R_2 = r$, and $d = 5$. So, we have: $|2 - r| < 5 < 2 + r$

This inequality can be split into two separate inequalities: a) $5 < 2 + r$ b) $|2 - r| < 5$

Let's solve each inequality:

a) From $5 < 2 + r$: $r > 5 - 2$ $r > 3$

b) From $|2 - r| < 5$: This inequality means $-5 < 2 - r < 5$. Subtract 2 from all parts of the inequality: $-5 - 2 < -r < 5 - 2$ $-7 < -r < 3$ Multiply by -1 and reverse the inequality signs: $-3 < r < 7$

Since radius $r$ must be a positive value, we consider $0 < r < 7$.

5. Combine the conditions for r: We need $r$ to satisfy both $r > 3$ and $0 < r < 7$. The intersection of these two intervals is $3 < r < 7$.

So, the set of all possible values of $r$ is the interval $(\alpha, \beta) = (3, 7)$. This means $\alpha = 3$ and $\beta = 7$.

6. Calculate the required expression: We need to find the value of $3\beta - 2\alpha$. Substitute the values of $\alpha$ and $\beta$: $3\beta - 2\alpha = 3(7) - 2(3)$ $= 21 - 6$ $= 15$

The final answer is $\boxed{\text{15}}$.