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Question:

Let C be the circle x² + y² + 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 60 degree. Then, the point at which L touches the line x = 6 is?

Explanation:

Let C be the circle x² + y² + 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 60 degree. Then, the point at which L touches the line x = 6 is?

Given, x² + y² + 4x - 6y - 3 = 0
⇒ x² + 4x + 4 + y² - 6y + 9 - 3 - 4 - 9 = 0 [Adding and Subtracting 4 and 9]
⇒ (x + 2)² + (y - 3)² = 16
⇒ (x + 2)² + (y - 3)² = 4² 

This represents the equation of a circle with radius = 4 units and center (O) at (-2, 3).

Point L lies on line x = 6, hence x-coordinate of L is 6. Let y-coordinate of L be 'y'.

Let P be the point where tangent from L touches the given circle.

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∆LOP is a 30-60-90° triangle.
⇒ OL = 2 × OP = 8

∴ OL = 8 = $\sqrt{(6 - {(-2))}^2+{(3-\mathrm{y})}^2}$
⇒ 64 = (8)² + (3 - y)² 
⇒ y = 3

∴ Coordinate of L = (6, 3)

Hence, option (c).

Note: A locus of points is a set of points that all satisfy some given condition or property.
Here L is the locus of the point of intersection of a pair of tangents to C with the angle between the two tangents equal to 60 degree. Hence, L will be a circle bigger than the original circle as shown. Tangents drawn from any point on this circle to the smaller circle will make an angle of 60 degree.