Write down the equation of the tangent to the circle
`4 x^(2) + 4y ^(2) - 16 x + 24 y = 117 ` at the point ( - 4, - `(11)/(2)` )

Answer

To find the equation of the tangent to the circle given by the equation \(4x^2 + 4y^2 - 16x + 24y = 117\) at the point \((-4, -\frac{11}{2})\), we will follow these steps:

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

\[ 4x^2 + 4y^2 - 16x + 24y - 117 = 0 \]

Dividing the entire equation by 4 gives us:

\[ x^2 + y^2 - 4x + 6y - \frac{117}{4} = 0 \]

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

For \(x\):
\[ x^2 - 4x \quad \text{can be written as} \quad (x - 2)^2 - 4 \]

For \(y\):
\[ y^2 + 6y \quad \text{can be written as} \quad (y + 3)^2 - 9 \]

Substituting these into the equation gives:

\[ (x - 2)^2 - 4 + (y + 3)^2 - 9 - \frac{117}{4} = 0 \]

Combining the constants:

\[ (x - 2)^2 + (y + 3)^2 - \left(4 + 9 + \frac{117}{4}\right) = 0 \]

Calculating \(4 + 9 = 13\) and converting to a common denominator:

\[ 13 = \frac{52}{4} \quad \Rightarrow \quad 4 + 9 + \frac{117}{4} = \frac{52 + 117}{4} = \frac{169}{4} \]

Thus, we have:

\[ (x - 2)^2 + (y + 3)^2 = \frac{169}{4} \]

This represents a circle with center \((2, -3)\) and radius \(\frac{13}{2}\).

### Step 3: Use the Tangent Formula
The formula for the tangent to a circle at a point \((x_1, y_1)\) is given by:

\[ xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c = 0 \]

Where \(g\), \(f\), and \(c\) are derived from the standard form of the circle. From our equation:

- \(g = -2\)
- \(f = 3\)
- \(c = -\frac{169}{4}\)

Substituting \(x_1 = -4\) and \(y_1 = -\frac{11}{2}\):

\[ x(-4) + y(-\frac{11}{2}) - 2(x - 4) + 3(y + \frac{11}{2}) - \frac{169}{4} = 0 \]

### Step 4: Simplify the Equation
Now we simplify:

\[ -4x - \frac{11}{2}y - 2x + 8 + 3y + \frac{33}{2} - \frac{169}{4} = 0 \]

Combining like terms:

\[ (-4x - 2x) + (-\frac{11}{2}y + 3y) + (8 + \frac{33}{2} - \frac{169}{4}) = 0 \]

Calculating the constants:

\[ 8 = \frac{32}{4}, \quad \frac{33}{2} = \frac{66}{4} \quad \Rightarrow \quad \frac{32 + 66 - 169}{4} = \frac{-71}{4} \]

Thus, the equation of the tangent simplifies to:

\[ -6x + \frac{y}{2} - \frac{71}{4} = 0 \]

### Final Equation
Multiplying through by 4 to eliminate the fraction gives:

\[ -24x + 2y - 71 = 0 \]

So, the equation of the tangent to the circle at the point \((-4, -\frac{11}{2})\) is:

\[ 24x - 2y + 71 = 0 \]