The locus of a moving point so that tangents from it to circle `x^(2)+y^(2)+4x-6y+9sin^(2)alpha+13cos^(2)alpha=0` are inclined at an angle `'2alpha'`Find the equation of the locus of the point . :

Options

A. `y^(2)=4(x+2)`
B. `x^(2)+y^(2)+4x-6y+9=0`
C. `x^(2)+y^(2)+4x-6y-9=0`
D. `3x^(2)+2y^(2)=12`

Explanation

To find the equation of the locus of a moving point such that the tangents from it to the given circle are inclined at an angle of \(2\alpha\), we will follow these steps:

### Step 1: Write the equation of the circle
The given equation of the circle is:
\[ x^2 + y^2 + 4x - 6y + 9\sin^2\alpha + 13\cos^2\alpha = 0 \]

### Step 2: Identify the center and radius of the circle
To find the center and radius, we rewrite the circle equation in standard form. We can complete the square for the \(x\) and \(y\) terms.

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

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

Now substituting back into the equation:
\[ (x + 2)^2 - 4 + (y - 3)^2 - 9 + 9\sin^2\alpha + 13\cos^2\alpha = 0 \]
\[ (x + 2)^2 + (y - 3)^2 + 9\sin^2\alpha + 13\cos^2\alpha - 13 = 0 \]
\[ (x + 2)^2 + (y - 3)^2 = 13 - 9\sin^2\alpha - 13\cos^2\alpha \]

The center of the circle is \((-2, 3)\) and the radius \(r\) is:
\[ r = \sqrt{13 - 9\sin^2\alpha - 13\cos^2\alpha} \]

### Step 3: Length of the tangent from point \(P(h, k)\)
The length of the tangent from point \(P(h, k)\) to the circle is given by:
\[ L = \sqrt{(h + 2)^2 + (k - 3)^2 - r^2} \]

Substituting the expression for \(r^2\):
\[ L = \sqrt{(h + 2)^2 + (k - 3)^2 - (13 - 9\sin^2\alpha - 13\cos^2\alpha)} \]

### Step 4: Using the angle condition
The angle between the tangents is given by:
\[ \tan(2\alpha) = \frac{2\tan(\alpha)}{1 - \tan^2(\alpha)} \]
Using the formula for the length of the tangent and the angle condition, we can relate the radius and the length of the tangent:
\[ \tan(\alpha) = \frac{r}{L} \]
Thus:
\[ \tan(2\alpha) = \frac{2r}{L^2} \]

### Step 5: Set up the equation
From the relation of the angles and the lengths, we can derive:
\[ L^2 = \frac{4r^2}{\tan^2(2\alpha)} \]

### Step 6: Substitute and simplify
Substituting the expressions for \(L\) and \(r\) into the equation, we can find the locus of the point \(P(h, k)\):
\[ (h + 2)^2 + (k - 3)^2 - (13 - 9\sin^2\alpha - 13\cos^2\alpha) = \frac{4(13 - 9\sin^2\alpha - 13\cos^2\alpha)}{\tan^2(2\alpha)} \]

### Final Result
After simplifying, we arrive at the equation of the locus of the point \(P(h, k)\):
\[ x^2 + y^2 + 4x - 6y + 9\sin^2\alpha + 13\cos^2\alpha - 4\cos^2\alpha = 0 \]