Question

Let \( A(1,2) \) be the centre and 3 be the radius of a circle \( S \). Let \( B(-1, -1) \) be the centre and \( r \) be the radius of another circle \( S_1 \). If \( \frac{\pi}{3} \) is the angle between the circles \( S \) and \( S_1 \), then the number of possible values of \( r \) is:

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

For two circles with a known angle between them, there can be two possible values for the radius of one circle depending on the configuration.

Answer 1

The Correct Option is B

We are given the centers and radii of two circles \( S \) and \( S_1 \). The angle between the two circles is \( \frac{\pi}{3} \). We need to find the number of possible values for the radius \( r \) of the second circle. From the geometry of the problem and applying the angle between two circles, there are exactly two possible values for the radius \( r \). Thus, the correct answer is option (2), 2.

Answer 2

Step 1: Understand the given parameters
We are given two circles:
- Circle \( S \) has center \( A(1, 2) \) and radius \( R = 3 \)
- Circle \( S_1 \) has center \( B(-1, -1) \) and radius \( r \)
- The angle between the two circles is \( \theta = \frac{\pi}{3} \)

Step 2: Use the formula for angle between two circles
The angle \( \theta \) between two intersecting circles with radii \( R \) and \( r \), and centers separated by a distance \( d \), is given by:
\[ \cos \theta = \frac{R^2 + d^2 - r^2}{2Rd} \]

Step 3: Compute the distance between centers
The distance \( d \) between centers \( A(1,2) \) and \( B(-1,-1) \) is:
\[ d = \sqrt{(1 + 1)^2 + (2 + 1)^2} = \sqrt{4 + 9} = \sqrt{13} \]

Step 4: Substitute values into the angle formula
Given \( \cos \frac{\pi}{3} = \frac{1}{2} \), \( R = 3 \), and \( d = \sqrt{13} \):
\[ \frac{1}{2} = \frac{9 + 13 - r^2}{2 \cdot 3 \cdot \sqrt{13}} = \frac{22 - r^2}{6\sqrt{13}} \]

Step 5: Solve the equation
Multiply both sides by \( 6\sqrt{13} \):
\[ 3\sqrt{13} = 22 - r^2 \Rightarrow r^2 = 22 - 3\sqrt{13} \]
This is one possibility. However, because the angle \( \frac{\pi}{3} \) can occur when the circles intersect from either side (inside or outside), we also consider the symmetric case with:
\[ \cos \theta = \frac{r^2 + d^2 - R^2}{2rd} \Rightarrow \frac{1}{2} = \frac{r^2 + 13 - 9}{2r\sqrt{13}} = \frac{r^2 + 4}{2r\sqrt{13}} \]
Now multiply both sides by \( 2r\sqrt{13} \):
\[ r\sqrt{13} = r^2 + 4 \Rightarrow r^2 - r\sqrt{13} + 4 = 0 \]
This quadratic in \( r \) will have two distinct real solutions if the discriminant is positive:
\[ \Delta = (\sqrt{13})^2 - 4 \cdot 1 \cdot 4 = 13 - 16 = -3 \Rightarrow \text{No real solution} \]
So, the only valid case is the original equation:
\[ r^2 = 22 - 3\sqrt{13} \quad \text{or} \quad r^2 = 22 + 3\sqrt{13} \]
Both yield real, positive values of \( r \), hence two values.

Final Answer:
2