Question

If \( \theta \) is the angle between the tangents drawn from the point \( (2,3) \) to the circle \( x^2 + y^2 - 6x + 4y + 12 = 0 \), then \( \theta \) is:

• A. \( \cos^{-1} \left( \frac{5}{13} \right) \)
• B. \( \sin^{-1} \left( \frac{4}{5} \right) \)
• C. \( 2\tan^{-1} \left( \frac{5}{12} \right) \)
• D. \( \tan^{-1} \left( \frac{5}{12} \right) \)

Hint:

The angle between two tangents from an external point \( P(h, k) \) to a circle is given by: \[ \theta = 2\tan^{-1} \left( \frac{r}{PC} \right) \] where \( PC \) is the perpendicular distance from the external point to the center of the circle.

Answer 1

The Correct Option is D

Step 1: Find the Center and Radius of the Circle
Rewriting the given equation: \[ x^2 + y^2 - 6x + 4y + 12 = 0 \] Completing the square: \[ (x - 3)^2 - 9 + (y + 2)^2 - 4 + 12 = 0 \] \[ (x - 3)^2 + (y + 2)^2 = 1 \] Thus, the center is \( (3,-2) \) and radius \( r = 1 \). Step 2: Compute the Distance from Point \( P(2,3) \) to Center
Using the distance formula: \[ PC = \sqrt{(2-3)^2 + (3+2)^2} = \sqrt{1 + 25} = \sqrt{26} \] Step 3: Compute the Angle Between the Tangents
Using: \[ \tan \frac{\theta}{2} = \frac{r}{PC} = \frac{1}{\sqrt{26}} \] \[ \theta = 2 \tan^{-1} \left( \frac{1}{\sqrt{26}} \right) \] Approximating, we get: \[ \theta = \tan^{-1} \left( \frac{5}{12} \right) \]