Find the equation of the tangent to
`x^(2) + y^(2) - 2x + 4y=0 " at" (3,-1)` Also find
the equation of tangent parallel to it.

Answer

To find the equation of the tangent to the circle given by the equation \( x^2 + y^2 - 2x + 4y = 0 \) at the point \( (3, -1) \), and also to find the equation of the tangent parallel to it, we can follow these steps:

### Step 1: Rewrite the Circle Equation
First, we rewrite the given equation of the circle in standard form. The equation is:

\[ x^2 + y^2 - 2x + 4y = 0 \]

We can complete the square for both \( x \) and \( y \).

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

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

Substituting these back into the equation gives:

\[ (x - 1)^2 - 1 + (y + 2)^2 - 4 = 0 \]

Simplifying this, we have:

\[ (x - 1)^2 + (y + 2)^2 = 5 \]

This shows that the center of the circle is \( (1, -2) \) and the radius is \( \sqrt{5} \).

### Step 2: Find the Slope of the Radius
Next, we find the slope of the radius to the point \( (3, -1) \). The slope \( m \) of the line connecting the center \( (1, -2) \) to the point \( (3, -1) \) is given by:

\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-1 - (-2)}{3 - 1} = \frac{1}{2} \]

### Step 3: Find the Slope of the Tangent
The slope of the tangent line is the negative reciprocal of the slope of the radius. Therefore, the slope \( m_t \) of the tangent is:

\[ m_t = -\frac{1}{m} = -2 \]

### Step 4: Use Point-Slope Form to Find the Tangent Equation
Using the point-slope form of the equation of a line \( y - y_1 = m(x - x_1) \), we substitute \( (x_1, y_1) = (3, -1) \) and \( m = -2 \):

\[ y - (-1) = -2(x - 3) \]

Simplifying this gives:

\[ y + 1 = -2x + 6 \]
\[ y = -2x + 5 \]

### Step 5: Find the Equation of the Parallel Tangent
Since parallel lines have the same slope, the equation of the tangent parallel to the one we just found will also have a slope of \( -2 \). The general form of the equation of a line with slope \( -2 \) can be written as:

\[ y = -2x + c \]

To find the value of \( c \) for the parallel tangent, we can choose any value for \( c \) that is different from \( 5 \) (the y-intercept of the original tangent). For example, if we choose \( c = -5 \):

\[ y = -2x - 5 \]

### Final Answer
The equation of the tangent to the circle at the point \( (3, -1) \) is:

\[ y = -2x + 5 \]

The equation of the tangent parallel to it is:

\[ y = -2x - 5 \]