The angle between a pair of tangents from a point P to the circe `x^2 + y^2+ 4 x-6y + 9 sin^2 alpha + 13 cos^2 alpha =0` is `2alpha`. Find the equation of the locus of the point P.

Answer

To find the equation of the locus of the point P from which tangents are drawn to the circle given by the equation \(x^2 + y^2 + 4x - 6y + 9 \sin^2 \alpha + 13 \cos^2 \alpha = 0\) and the angle between the tangents is \(2\alpha\), we can follow these steps:

### Step 1: Identify the Circle's Center and Radius
The given equation of the circle can be rewritten in the standard form \(x^2 + y^2 + 2gx + 2fy + c = 0\). From the equation:

\[ x^2 + y^2 + 4x - 6y + (9 \sin^2 \alpha + 13 \cos^2 \alpha) = 0 \]

we can identify:
- \(2g = 4 \Rightarrow g = 2\)
- \(2f = -6 \Rightarrow f = -3\)
- \(c = 9 \sin^2 \alpha + 13 \cos^2 \alpha\)

Thus, the center of the circle \(C\) is at \((-g, -f) = (-2, 3)\).

### Step 2: Calculate the Radius
The radius \(r\) of the circle can be calculated using the formula:

\[ r = \sqrt{g^2 + f^2 - c} \]

Substituting the values we found:

\[ r = \sqrt{2^2 + (-3)^2 - (9 \sin^2 \alpha + 13 \cos^2 \alpha)} \]

Calculating \(g^2 + f^2\):

\[ g^2 + f^2 = 4 + 9 = 13 \]

Thus,

\[ r = \sqrt{13 - (9 \sin^2 \alpha + 13 \cos^2 \alpha)} \]

### Step 3: Simplify the Radius Expression
We can simplify the expression for the radius:

\[ r = \sqrt{13 - 9 \sin^2 \alpha - 13 \cos^2 \alpha} \]

Using the identity \(\sin^2 \alpha + \cos^2 \alpha = 1\):

\[ r = \sqrt{13 - 9(1 - \cos^2 \alpha) - 13 \cos^2 \alpha} \]
\[ = \sqrt{13 - 9 + 9 \cos^2 \alpha - 13 \cos^2 \alpha} \]
\[ = \sqrt{4 - 4 \cos^2 \alpha} \]
\[ = 2 \sqrt{1 - \cos^2 \alpha} = 2 \sin \alpha \]

### Step 4: Use the Tangent Length Formula
The length of the tangents from point \(P(h, k)\) to the circle is given by:

\[ PT^2 = PC^2 - r^2 \]

Where \(PC\) is the distance from point \(P\) to the center \(C(-2, 3)\):

\[ PC = \sqrt{(h + 2)^2 + (k - 3)^2} \]

Thus, we have:

\[ PT^2 = (h + 2)^2 + (k - 3)^2 - (2 \sin \alpha)^2 \]
\[ = (h + 2)^2 + (k - 3)^2 - 4 \sin^2 \alpha \]

### Step 5: Relate to the Angle Between Tangents
Given that the angle between the tangents is \(2\alpha\), we know that:

\[ \frac{PT}{PC} = \sin \alpha \]

Squaring both sides gives:

\[ \frac{PT^2}{PC^2} = \sin^2 \alpha \]

Substituting \(PT^2\) and \(PC^2\):

\[ \frac{(h + 2)^2 + (k - 3)^2 - 4 \sin^2 \alpha}{(h + 2)^2 + (k - 3)^2} = \sin^2 \alpha \]

### Step 6: Rearranging the Equation
Cross-multiplying gives:

\[ (h + 2)^2 + (k - 3)^2 - 4 \sin^2 \alpha = \sin^2 \alpha \cdot ((h + 2)^2 + (k - 3)^2) \]

Rearranging leads to:

\[ (h + 2)^2 + (k - 3)^2 (1 - \sin^2 \alpha) = 4 \sin^2 \alpha \]

### Step 7: Final Locus Equation
Using \(\sin^2 \alpha + \cos^2 \alpha = 1\):

\[ (h + 2)^2 + (k - 3)^2 = 4 \]

Substituting \(h\) and \(k\) back to \(x\) and \(y\):

\[ (x + 2)^2 + (y - 3)^2 = 4 \]

This represents a circle centered at \((-2, 3)\) with a radius of \(2\).

### Final Answer
The equation of the locus of point \(P\) is:

\[ (x + 2)^2 + (y - 3)^2 = 4 \]