Find the equation of the tangents to the circle `x^(2)+y^(2)-4x+6y-12=0` which are parallel to `x+y-8=0`

Answer

To find the equation of the tangents to the circle \(x^2 + y^2 - 4x + 6y - 12 = 0\) that are parallel to the line \(x + y - 8 = 0\), we will follow these steps:

### Step 1: Rewrite the equation of the circle in standard form
The given equation of the circle is:
\[ x^2 + y^2 - 4x + 6y - 12 = 0 \]
We can rearrange this equation by completing the square for both \(x\) and \(y\).

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

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

Substituting these back into the circle's equation:
\[ (x - 2)^2 - 4 + (y + 3)^2 - 9 - 12 = 0 \]
\[ (x - 2)^2 + (y + 3)^2 - 25 = 0 \]
Thus, the standard form of the circle is:
\[ (x - 2)^2 + (y + 3)^2 = 25 \]

### Step 2: Identify the center and radius of the circle
From the standard form \((x - h)^2 + (y - k)^2 = r^2\):
- Center \((h, k) = (2, -3)\)
- Radius \(r = \sqrt{25} = 5\)

### Step 3: Find the slope of the given line
The equation of the line is:
\[ x + y - 8 = 0 \]
This can be rewritten as:
\[ y = -x + 8 \]
The slope \(m\) of this line is \(-1\).

### Step 4: Use the point-slope form to find the tangent lines
The equation of the tangent line to the circle can be expressed as:
\[ y - k = m(x - h) + r\sqrt{1 + m^2} \]
Substituting \(m = -1\), \(h = 2\), \(k = -3\), and \(r = 5\):
\[ y + 3 = -1(x - 2) \pm 5\sqrt{1 + (-1)^2} \]
Calculating \(\sqrt{1 + 1} = \sqrt{2}\):
\[ y + 3 = -1(x - 2) \pm 5\sqrt{2} \]
This simplifies to:
\[ y + 3 = -x + 2 \pm 5\sqrt{2} \]

### Step 5: Rearranging to find the equations of the tangents
1. For the positive case:
\[ y = -x + 2 + 5\sqrt{2} - 3 \]
\[ y = -x - 1 + 5\sqrt{2} \]
\[ x + y + 1 - 5\sqrt{2} = 0 \]

2. For the negative case:
\[ y = -x + 2 - 5\sqrt{2} - 3 \]
\[ y = -x - 1 - 5\sqrt{2} \]
\[ x + y + 1 + 5\sqrt{2} = 0 \]

### Final Answer
The equations of the tangents to the circle that are parallel to the line \(x + y - 8 = 0\) are:
\[ x + y + 1 - 5\sqrt{2} = 0 \quad \text{and} \quad x + y + 1 + 5\sqrt{2} = 0 \]