The equation of the normal to the circle `x^(2)+y^(2)+6x+4y-3=0` at (1,-2) to is

Options

A. `y+1=0`
B. `y+2=0`
C. `y+3=0`
D. `y-2=0`

Answer

To find the equation of the normal to the circle given by the equation \( x^2 + y^2 + 6x + 4y - 3 = 0 \) at the point \( (1, -2) \), we can follow these steps:

### Step 1: Rewrite the Circle Equation
First, we need to rewrite the equation of the circle in standard form. The given equation is:
\[ x^2 + y^2 + 6x + 4y - 3 = 0 \]
We can rearrange it as:
\[ x^2 + 6x + y^2 + 4y = 3 \]

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

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

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

Substituting these back into the equation gives:
\[ (x + 3)^2 - 9 + (y + 2)^2 - 4 = 3 \]
This simplifies to:
\[ (x + 3)^2 + (y + 2)^2 = 16 \]

### Step 3: Identify the Center and Radius
From the standard form \( (x + 3)^2 + (y + 2)^2 = 16 \), we can identify:
- Center \( C(-3, -2) \)
- Radius \( r = 4 \)

### Step 4: Find the Slope of the Radius
The radius at the point \( (1, -2) \) is the line connecting the center \( C(-3, -2) \) to the point \( (1, -2) \). The slope \( m \) of the radius is calculated as follows:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-2 - (-2)}{1 - (-3)} = \frac{0}{4} = 0 \]

### Step 5: Find the Slope of the Normal
The normal line is perpendicular to the radius. Since the slope of the radius is \( 0 \) (horizontal line), the slope of the normal will be undefined, indicating a vertical line.

### Step 6: Write the Equation of the Normal
A vertical line through the point \( (1, -2) \) can be expressed as:
\[ x = 1 \]

### Conclusion
Thus, the equation of the normal to the circle at the point \( (1, -2) \) is:
\[ \boxed{x = 1} \]