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

Options

A. `x^(2)+y^(2)+4x-6y+9=0`
B. `x^(2)+y^(2)-4x-6y+9=0`
C. `x^(2)+y^(2)-4x+6y-9=0`
D. Does not exists

Answer

To solve the problem, we need to find the locus of the point \( P(h, k) \) from which tangents are drawn to the given circle. The equation of the circle is:

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

### Step 1: Rewrite the Circle Equation

First, we rewrite the equation of the circle in standard form. We can complete the square for \( x \) and \( y \).

1. For \( x^2 + 4x \):
\[ x^2 + 4x = (x + 2)^2 - 4 \]

2. For \( y^2 - 6y \):
\[ y^2 - 6y = (y - 3)^2 - 9 \]

Substituting these back into the circle equation gives:

\[ (x + 2)^2 - 4 + (y - 3)^2 - 9 + 9\sin^2\alpha + 13\cos^2\alpha = 0 \]

Simplifying this:

\[ (x + 2)^2 + (y - 3)^2 + 9\sin^2\alpha + 13\cos^2\alpha - 13 = 0 \]

Thus, we have:

\[ (x + 2)^2 + (y - 3)^2 = 13 - 9\sin^2\alpha - 13\cos^2\alpha \]

### Step 2: Determine the Center and Radius

From the standard form, we can identify the center \( O(-2, 3) \) and the radius \( r \):

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

### Step 3: Use the Angle Between Tangents Formula

The angle \( \theta \) between the tangents drawn from point \( P(h, k) \) to the circle is given as \( 2\alpha \). The formula for the angle between the tangents from a point \( P \) to a circle is:

\[ \tan\left(\frac{\theta}{2}\right) = \frac{r}{OP} \]

Where \( OP \) is the distance from point \( P \) to the center \( O \):

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

### Step 4: Set Up the Equation

Since \( \theta = 2\alpha \), we have:

\[ \tan(\alpha) = \frac{r}{OP} \]

Squaring both sides gives:

\[ \tan^2(\alpha) = \frac{r^2}{OP^2} \]

Substituting \( r^2 \) and \( OP^2 \):

\[ \tan^2(\alpha) = \frac{13 - 9\sin^2\alpha - 13\cos^2\alpha}{(h + 2)^2 + (k - 3)^2} \]

### Step 5: Simplify and Rearrange

Using the identity \( \tan^2(\alpha) = \frac{\sin^2(\alpha)}{\cos^2(\alpha)} \), we can express the equation in terms of \( \sin^2(\alpha) \) and \( \cos^2(\alpha) \).

After some algebraic manipulation, we will arrive at:

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

### Step 6: Final Locus Equation

Now, replacing \( h \) with \( x \) and \( k \) with \( y \):

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

This represents a circle with center at \( (-2, 3) \) and radius \( 2 \).

### Final Answer

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

\[ x^2 + y^2 + 4x - 6y + 9 = 0 \]