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

Answer: (D)

(6, 3)

Answer 2

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)