Find the equation of the tangent to the circle `x^2 + y^2-30x+6y+109=0` at `(4, -1)`

Options

A. `11x-2y-46=0`
B. `11x-3y-47=0`
C. `10x-3y-43=0`
D. `11x+2y-42=0`

Explanation

To find the equation of the tangent to the circle given by the equation \(x^2 + y^2 - 30x + 6y + 109 = 0\) at the point \((4, -1)\), we will 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:

\[ x^2 + y^2 - 30x + 6y + 109 = 0 \]

We can rearrange it as:

\[ x^2 - 30x + y^2 + 6y + 109 = 0 \]

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

For \(x\):
\[ x^2 - 30x = (x - 15)^2 - 225 \]

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

Substituting these back into the equation:

\[ (x - 15)^2 - 225 + (y + 3)^2 - 9 + 109 = 0 \]

Simplifying this gives:

\[ (x - 15)^2 + (y + 3)^2 - 225 - 9 + 109 = 0 \]
\[ (x - 15)^2 + (y + 3)^2 - 125 = 0 \]
\[ (x - 15)^2 + (y + 3)^2 = 125 \]

This shows that the circle has center \((15, -3)\) and radius \(\sqrt{125} = 5\sqrt{5}\).

### Step 3: Use the Point Form of the Tangent Equation
The point form of the tangent to a circle at point \((x_1, y_1)\) is given by:

\[ x_1x + y_1y + g(x + x_1) + f(y + y_1) + c = 0 \]

Where \(g\), \(f\), and \(c\) are derived from the standard form of the circle \(x^2 + y^2 + 2gx + 2fy + c = 0\).

From our circle equation, we have:
- \(g = -15\)
- \(f = 3\)
- \(c = -125\)

### Step 4: Substitute the Point \((4, -1)\)
Now, substituting \(x_1 = 4\) and \(y_1 = -1\):

\[ 4x + (-1)y - 15(x + 4) + 3(y - 1) - 125 = 0 \]

Expanding this gives:

\[ 4x - y - 15x - 60 + 3y - 3 - 125 = 0 \]
\[ -11x + 2y - 188 = 0 \]

### Step 5: Rearranging the Equation
Rearranging the equation gives:

\[ 11x - 2y - 188 = 0 \]

### Final Equation
Thus, the equation of the tangent to the circle at the point \((4, -1)\) is:

\[ 11x - 2y - 188 = 0 \]