Question

If \( \theta \) is the angle between the circles \( x^2+y^2-4x+2y-4=0 \) and \( x^2+y^2-2x+4y-11=0 \), then \( \sin\theta = \)

• A. \( \frac{\sqrt{47}}{24} \)
• B. \( \frac{23}{25} \)
• C. \( \frac{23}{24} \)
• D. \( \frac{\sqrt{3}}{5} \)

Hint:

The angle between two intersecting circles is defined as the angle between their tangents at a point of intersection. This angle can be found directly from the radii and the distance between centers using the cosine rule on the triangle formed by the centers and an intersection point.

Answer 1

The Correct Option is A

Step 1: Find the centers and radii of both circles.
For the first circle, S1: \(x^2+y^2-4x+2y-4=0\).
Center \(C_1 = (2, -1)\).
Radius \(r_1 = \sqrt{2^2 + (-1)^2 - (-4)} = \sqrt{4+1+4} = \sqrt{9} = 3\).
For the second circle, S2: \(x^2+y^2-2x+4y-11=0\).
Center \(C_2 = (1, -2)\).
Radius \(r_2 = \sqrt{1^2 + (-2)^2 - (-11)} = \sqrt{1+4+11} = \sqrt{16} = 4\).
Step 2: Find the distance between the centers.
Let \(d\) be the distance between \(C_1\) and \(C_2\).
\(d = \sqrt{(2-1)^2 + (-1 - (-2))^2} = \sqrt{1^2 + 1^2} = \sqrt{2}\).
Step 3: Use the formula for the angle between two circles.
The angle \( \theta \) is given by the formula \( \cos\theta = \frac{r_1^2 + r_2^2 - d^2}{2r_1r_2} \).
\( \cos\theta = \frac{3^2 + 4^2 - (\sqrt{2})^2}{2 \cdot 3 \cdot 4} = \frac{9 + 16 - 2}{24} = \frac{23}{24} \).
Step 4: Find \( \sin\theta \).
Using the identity \( \sin^2\theta + \cos^2\theta = 1 \). Since the angle between circles is usually taken to be acute, \( \sin\theta \) will be positive.
\( \sin^2\theta = 1 - \cos^2\theta = 1 - \left(\frac{23}{24}\right)^2 = 1 - \frac{529}{576} = \frac{576-529}{576} = \frac{47}{576} \).
\( \sin\theta = \sqrt{\frac{47}{576}} = \frac{\sqrt{47}}{24} \).