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 $$ 60^\circ $$. Then, the point at which $$ L $$ touches the line $$ x = 6 $$ is:

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Accepted Answer

#### Step 1: Rewrite the circle equation
The given circle equation is:  
$$ x^2 + y^2 + 4x - 6y - 3 = 0. $$  
Rewriting it in standard form:  
$$ (x + 2)^2 + (y - 3)^2 = 16. $$  
This represents a circle with center at $$ (-2, 3) $$ and radius $$ r = 4 $$.

#### Step 2: Locus of points w![Screenshot 2025-03-30 20.20.23.png](https://res.cloudinary.com/aptidude/image/upload/v1743346237/aptidude_questions/quyc98aftasqq4e5ycuh.png)here tangents make a $$ 60^\circ $$ angle
From a point $$ P(h, k) $$, two tangents are drawn to the circle such that the angle between them is $$ 60^\circ $$. Using the formula for the angle between tangents:  
$$ \sin\left(\frac{	heta}{2}ight) = \frac{r}{d}, $$  
where $$ d $$ is the distance from $$ P(h, k) $$ to the center of the circle and radius is given as $$ r = 4 $$. For an angle of $$ 	heta = 60^\circ $$:  
$$ \sin(30^\circ) = \frac{4}{d}. $$  
Since $$ \sin(30^\circ) = \frac{1}{2} $$, we get:  
$$ d = 8. $$

Thus, the locus of point $$ P(h, k) $$ is a circle centered at $$ (-2, 3) $$ with radius equal to **8**.

#### Step 3: Equation of locus
The equation of this locus is:  
$$ (x + 2)^2 + (y - 3)^2 = 64. $$

#### Step 4: Point where locus touches line $$ x = 6 $$
Substitute $$ x = 6 $$ into the locus equation:  
$$ (6 + 2)^2 + (y - 3)^2 = 64. $$
Simplify:  
$$ (8)^2 + (y - 3)^2 = 64. $$
$$ 64 + (y - 3)^2 = 64. $$
$$ (y - 3)^2 = 0. $$
Thus,  
$$ y - 3 = 0 \quad 	ext{or} \quad y = 3. $$
The point of intersection is **(6, 3)**.

#### Final Answer:
The point at which the locus touches the line $$ x = 6 $$ is **(6, 3)**.

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