Find the equation of tangents of the circle
`x^(2) + y^(2) - 8 x - 2y + 12 = 0 ` at the points
whose ordinates are -1.

Answer

To find the equations of the tangents to the circle given by the equation \( x^2 + y^2 - 8x - 2y + 12 = 0 \) at the points where the ordinates are -1, we can follow these steps:

### Step 1: Rewrite the Circle Equation
First, we will rewrite the circle equation in standard form by completing the square.

The given equation is:
\[ x^2 + y^2 - 8x - 2y + 12 = 0 \]
Rearranging gives:
\[ x^2 - 8x + y^2 - 2y + 12 = 0 \]

### Step 2: Complete the Square
Now, we complete the square for \(x\) and \(y\).

For \(x\):
\[ x^2 - 8x = (x - 4)^2 - 16 \]

For \(y\):
\[ y^2 - 2y = (y - 1)^2 - 1 \]

Substituting back into the equation:
\[ (x - 4)^2 - 16 + (y - 1)^2 - 1 + 12 = 0 \]
Simplifying gives:
\[ (x - 4)^2 + (y - 1)^2 - 5 = 0 \]
Thus, we can write the equation of the circle as:
\[ (x - 4)^2 + (y - 1)^2 = 5 \]
This means the center of the circle is at \( (4, 1) \) and the radius is \( \sqrt{5} \).

### Step 3: Find Points on the Circle with Ordinate -1
We need to find the points on the circle where the ordinate (y-coordinate) is -1. Setting \(y = -1\) in the circle's equation:
\[ (x - 4)^2 + (-1 - 1)^2 = 5 \]
This simplifies to:
\[ (x - 4)^2 + (-2)^2 = 5 \]
\[ (x - 4)^2 + 4 = 5 \]
\[ (x - 4)^2 = 1 \]
Taking the square root gives:
\[ x - 4 = 1 \quad \text{or} \quad x - 4 = -1 \]
So,
\[ x = 5 \quad \text{or} \quad x = 3 \]
Thus, the points are \( (5, -1) \) and \( (3, -1) \).

### Step 4: Find the Slope of the Tangents
Next, we need to find the slope of the tangent lines at these points. We will differentiate the circle's equation implicitly.

Differentiating \(x^2 + y^2 - 8x - 2y + 12 = 0\):
\[ 2x + 2y \frac{dy}{dx} - 8 - 2\frac{dy}{dx} = 0 \]
Rearranging gives:
\[ (2y - 2) \frac{dy}{dx} = 8 - 2x \]
Thus,
\[ \frac{dy}{dx} = \frac{8 - 2x}{2y - 2} = \frac{4 - x}{y - 1} \]

### Step 5: Calculate the Slopes at the Points
1. For the point \( (5, -1) \):
\[ \frac{dy}{dx} = \frac{4 - 5}{-1 - 1} = \frac{-1}{-2} = \frac{1}{2} \]

2. For the point \( (3, -1) \):
\[ \frac{dy}{dx} = \frac{4 - 3}{-1 - 1} = \frac{1}{-2} = -\frac{1}{2} \]

### Step 6: Write the Equations of the Tangents
Using the point-slope form of the line \(y - y_1 = m(x - x_1)\):

1. For the point \( (5, -1) \) with slope \( \frac{1}{2} \):
\[ y + 1 = \frac{1}{2}(x - 5) \]
Multiplying through by 2:
\[ 2y + 2 = x - 5 \implies x - 2y = 7 \]

2. For the point \( (3, -1) \) with slope \( -\frac{1}{2} \):
\[ y + 1 = -\frac{1}{2}(x - 3) \]
Multiplying through by 2:
\[ 2y + 2 = -x + 3 \implies x + 2y = 1 \]

### Final Answer
The equations of the tangents are:
1. \( x - 2y = 7 \)
2. \( x + 2y = 1 \)