Find the equation of the tangent at the
point `30^(@)` (parametric value of `theta`) of the
circle is `x^(2) + y^(2) + 4x +6y - 39=0`.

Answer

To find the equation of the tangent at the point corresponding to the parametric angle \( \theta = 30^\circ \) of the given circle, we can follow these steps:

### Step 1: Rewrite the Circle Equation
The given equation of the circle is:
\[ x^2 + y^2 + 4x + 6y - 39 = 0 \]
We can rewrite this in standard form by completing the square.

### Step 2: Complete the Square
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 equation:
\[ (x + 2)^2 - 4 + (y + 3)^2 - 9 - 39 = 0 \]
This simplifies to:
\[ (x + 2)^2 + (y + 3)^2 - 52 = 0 \]
Thus, the standard form of the circle is:
\[ (x + 2)^2 + (y + 3)^2 = 52 \]

### Step 3: Identify the Center and Radius
From the standard form, we can identify:
- Center \( (-2, -3) \)
- Radius \( r = \sqrt{52} = 2\sqrt{13} \)

### Step 4: Find the Point on the Circle for \( \theta = 30^\circ \)
Using the parametric equations for the circle:
\[ x = h + r \cos(\theta) \quad \text{and} \quad y = k + r \sin(\theta) \]
where \( (h, k) \) is the center of the circle.

Substituting \( h = -2 \), \( k = -3 \), \( r = 2\sqrt{13} \), and \( \theta = 30^\circ \):
\[ x = -2 + 2\sqrt{13} \cos(30^\circ) = -2 + 2\sqrt{13} \cdot \frac{\sqrt{3}}{2} = -2 + \sqrt{39} \]
\[ y = -3 + 2\sqrt{13} \sin(30^\circ) = -3 + 2\sqrt{13} \cdot \frac{1}{2} = -3 + \sqrt{13} \]

### Step 5: Write the Tangent Equation
The equation of the tangent line at a point \( (x_0, y_0) \) on the circle is given by:
\[ (x + 2)(x_0 + 2) + (y + 3)(y_0 + 3) = 52 \]
Substituting \( (x_0, y_0) = (\sqrt{39} - 2, \sqrt{13} - 3) \):
\[ (x + 2)(\sqrt{39}) + (y + 3)(\sqrt{13}) = 52 \]

### Step 6: Simplify the Tangent Equation
Expanding and simplifying:
\[ \sqrt{39}x + 2\sqrt{39} + \sqrt{13}y + 3\sqrt{13} = 52 \]
Rearranging gives:
\[ \sqrt{39}x + \sqrt{13}y + (2\sqrt{39} + 3\sqrt{13} - 52) = 0 \]

### Final Tangent Equation
Thus, the equation of the tangent line at the point corresponding to \( \theta = 30^\circ \) is:
\[ \sqrt{39}x + \sqrt{13}y + (2\sqrt{39} + 3\sqrt{13} - 52) = 0 \]