Question

Let C be the circle \(x^2+y^2+4x-6y-3=0\) and \(L\) be the locus of the point of intersection of a pair of tangents to \(C\) with the angle between the two tangents equal to \(60º\) . Then, the point at which \(L\) touches the line \(x = 6\) is

• A. (6, 6)
• B. (6, 8)
• C. (6, 4)
• D. (6, 3)

Answer 1

The Correct Option is D

Provided:
Circle Equation: x² + y² + 4x - 6y - 3 = 0
Circle Radius: \( \sqrt{g^2+f^2-c} \)
\( =\sqrt{4+9+3}=\sqrt{16} \)
= 4
Circle Center: (-2, 3)
Let the point of intersection of tangents be (h, k).
The line joining (h, k) to the center forms a 30° angle with the tangent. Therefore, sin(30°) equals the ratio of the radius to the distance between the center and (h, k).
⇒ sin (30) = \( \frac{4}{\sqrt{(h-(-2))^2+(k-3)^2}} \)
⇒ sin (30) = \( \frac{4}{\sqrt{(h+2)^2+(k-3)^2}} \)
Squaring both sides yields:
\( \frac{1}{4} = \frac{16}{(h+2)^2+(k-3)^2} \)
⇒ (h + 2)² + (k - 3)² = 64
Given x = 6, implying h = 6:
⇒ (6 + 2)² + (k - 3)² = 64
⇒ 64 + (k - 3)² = 64
⇒ (k - 3)² = 0
k = 3.
Thus, the required point is (6, 3).
The correct option is (D): (6, 3).