Find the equation of the tangent to the circle ` x^(2) + y^(2) + 5x - 3y - 4 = 0 ` at the point `(1,2)` .

Options

A. `x +7 y = 9`
B. `x + y = 9`
C. `7x + y = 9`
D. `x +7 y = 12`

Explanation

To find the equation of the tangent to the circle given by the equation \( x^2 + y^2 + 5x - 3y - 4 = 0 \) at the point \( (1, 2) \), we will follow these steps:

### Step 1: Identify the center and radius of the circle
The general form of a circle's equation is \( x^2 + y^2 + 2gx + 2fy + c = 0 \).

From the given equation:
- \( 2g = 5 \) → \( g = \frac{5}{2} \)
- \( 2f = -3 \) → \( f = -\frac{3}{2} \)

The center \( C \) of the circle can be found using the coordinates \( (-g, -f) \):
- Center \( C = \left(-\frac{5}{2}, \frac{3}{2}\right) \)

### Step 2: Find the slope of the line connecting the center to the point of tangency
Let the point of tangency be \( P(1, 2) \).

The slope \( m_{CP} \) of the line segment \( CP \) is given by:
\[ m_{CP} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - \frac{3}{2}}{1 - \left(-\frac{5}{2}\right)} \]
Calculating this:
\[ m_{CP} = \frac{2 - 1.5}{1 + 2.5} = \frac{0.5}{3} = \frac{1}{6} \]

### Step 3: Find the slope of the tangent line
The slope of the tangent line \( m_{T} \) at point \( P \) is the negative reciprocal of the slope of \( CP \):
\[ m_{T} = -\frac{1}{m_{CP}} = -6 \]

### Step 4: Use the point-slope form to find the equation of the tangent line
Using the point-slope form of the line equation \( y - y_1 = m(x - x_1) \):
\[ y - 2 = -6(x - 1) \]
Expanding this:
\[ y - 2 = -6x + 6 \]
\[ y = -6x + 8 \]

### Final Answer
The equation of the tangent to the circle at the point \( (1, 2) \) is:
\[ y = -6x + 8 \]