Find the equations of the tangent to the
circle `x^(2) + y^(2) + 2x - 2y -3= 0` which are
perpendicular to `3x - y + 4 = 0`

Answer

To find the equations of the tangents to the circle \(x^2 + y^2 + 2x - 2y - 3 = 0\) that are perpendicular to the line \(3x - y + 4 = 0\), we can follow these steps:

### Step 1: Rewrite the Circle's Equation
The given equation of the circle is:
\[ x^2 + y^2 + 2x - 2y - 3 = 0 \]
We can rearrange this into the standard form of a circle by completing the square.

### Step 2: Complete the Square
1. For \(x^2 + 2x\), we complete the square:
\[ x^2 + 2x = (x + 1)^2 - 1 \]
2. For \(y^2 - 2y\), we complete the square:
\[ y^2 - 2y = (y - 1)^2 - 1 \]
3. Substitute these back into the circle's equation:
\[ (x + 1)^2 - 1 + (y - 1)^2 - 1 - 3 = 0 \]
Simplifying gives:
\[ (x + 1)^2 + (y - 1)^2 - 5 = 0 \implies (x + 1)^2 + (y - 1)^2 = 5 \]
This shows that the center of the circle is \((-1, 1)\) and the radius \(r = \sqrt{5}\).

### Step 3: Find the Slope of the Given Line
The line \(3x - y + 4 = 0\) can be rewritten in slope-intercept form:
\[ y = 3x + 4 \]
The slope \(m_1\) of this line is \(3\).

### Step 4: Determine the Slope of the Tangent
Since the tangents we are looking for are perpendicular to this line, the slope \(m_2\) of the tangent will be:
\[ m_2 = -\frac{1}{m_1} = -\frac{1}{3} \]

### Step 5: Use the Point-Slope Form of the Tangent Line
The equation of the tangent line at a point \((x_1, y_1)\) on the circle can be expressed as:
\[ y - y_1 = m_2(x - x_1) \]
Here, \((x_1, y_1)\) is a point on the circle, and \(m_2 = -\frac{1}{3}\).

### Step 6: Find the Points of Tangency
We can find the points where the tangent lines touch the circle. The general equation of the tangent line can be written as:
\[ y - 1 = -\frac{1}{3}(x + 1) \]
Rearranging gives:
\[ y = -\frac{1}{3}x + \frac{2}{3} \]

### Step 7: Substitute into the Circle's Equation
Substituting \(y = -\frac{1}{3}x + \frac{2}{3}\) into the circle's equation:
\[ (x + 1)^2 + \left(-\frac{1}{3}x + \frac{2}{3} - 1\right)^2 = 5 \]
This simplifies to:
\[ (x + 1)^2 + \left(-\frac{1}{3}x - \frac{1}{3}\right)^2 = 5 \]
Expanding and simplifying will yield a quadratic equation in \(x\).

### Step 8: Solve the Quadratic Equation
We will solve this quadratic equation to find the \(x\)-coordinates of the points of tangency. After finding \(x\), substitute back to find \(y\).

### Step 9: Write the Final Tangent Equations
Using the points of tangency found, substitute back into the tangent line equation to find the final equations of the tangents.