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

Given :
Equation of circle = x² + y² + 4x - 6y - 3 = 0
Radius of the circle = \(\sqrt{g^2+f^2-c}\)
\(=\sqrt{4+9+3}=\sqrt{16}\)
= 4
Center of the circle : (2, -3)
Suppose the point of intersection of the tangents is (h, k)
Now, the angle created by the line joining (h, k) to the centre makes an angle of 30° with the tangent and sin(30) will be the ratio of radius and distance between the center and (h, k)
⇒ sin (30) = \(\frac{4}{\sqrt{(h+2)^2+(k-3)^2}}\)
Now, by squaring on both sides , we get :
\(\frac{1}{4}=\frac{16}{(h+2)^2+(k-3)^2}\)
⇒ (h + 2)² + (k - 3)² = 64
Now , when x = 6 ⇒ h = 6 , we get
⇒ (6 + 2)² + (k - 3)² = 64
⇒ 64 + (k - 3)² = 64
k = 3.
Therefore, the required point is (6, 3)
So, the correct option is (D) : (6, 3)