Find the equation of tangent and normal at
`(3, 2)` of the circle `x^(2) + y^(2) - 3y - 4 = 0. `

Answer

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

### Step 1: Rewrite the circle equation
First, we need to rewrite the circle equation in standard form. The given equation is:
\[ x^2 + y^2 - 3y - 4 = 0 \]
We can rearrange it as:
\[ x^2 + y^2 - 3y = 4 \]
Next, we complete the square for the \(y\) terms.

### Step 2: Complete the square
To complete the square for \(y^2 - 3y\):
\[ y^2 - 3y = (y - \frac{3}{2})^2 - \frac{9}{4} \]
Substituting this back into the equation gives:
\[ x^2 + \left(y - \frac{3}{2}\right)^2 - \frac{9}{4} = 4 \]
Adding \(\frac{9}{4}\) to both sides:
\[ x^2 + \left(y - \frac{3}{2}\right)^2 = 4 + \frac{9}{4} = \frac{25}{4} \]
Thus, the standard form of the circle is:
\[ \left(x - 0\right)^2 + \left(y - \frac{3}{2}\right)^2 = \left(\frac{5}{2}\right)^2 \]
This shows that the center of the circle is at \((0, \frac{3}{2})\) and the radius is \(\frac{5}{2}\).

### Step 3: Find the slope of the radius
The slope of the radius from the center \((0, \frac{3}{2})\) to the point \((3, 2)\) is calculated as follows:
\[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - \frac{3}{2}}{3 - 0} = \frac{\frac{1}{2}}{3} = \frac{1}{6} \]

### Step 4: Find the slope of the tangent
The slope of the tangent line is the negative reciprocal of the slope of the radius:
\[ \text{slope of tangent} = -\frac{1}{\frac{1}{6}} = -6 \]

### Step 5: Write the equation of the tangent
Using the point-slope form of the line equation \(y - y_1 = m(x - x_1)\), where \(m\) is the slope and \((x_1, y_1)\) is the point (3, 2):
\[ y - 2 = -6(x - 3) \]
Expanding this:
\[ y - 2 = -6x + 18 \]
Rearranging gives:
\[ 6x + y - 20 = 0 \]
Thus, the equation of the tangent is:
\[ 6x + y - 20 = 0 \]

### Step 6: Find the slope of the normal
The slope of the normal line is simply the slope of the radius, which we found to be \(\frac{1}{6}\).

### Step 7: Write the equation of the normal
Using the point-slope form again:
\[ y - 2 = \frac{1}{6}(x - 3) \]
Expanding this:
\[ y - 2 = \frac{1}{6}x - \frac{1}{2} \]
Rearranging gives:
\[ 6y - 12 = x - 3 \]
Thus:
\[ x - 6y + 9 = 0 \]
So, the equation of the normal is:
\[ x - 6y + 9 = 0 \]

### Final Answers:
- Equation of the tangent: \(6x + y - 20 = 0\)
- Equation of the normal: \(x - 6y + 9 = 0\)