The equation of normal to the circle
`x^(2)+y^(2)- 4x+4y - 17=0`
which passes through (1, 1) is

Options

A. `3x + y - 4=0`
B. `x-y=0`
C. `x+y=0 `
D. none

Answer

To find the equation of the normal to the circle given by the equation \(x^2 + y^2 - 4x + 4y - 17 = 0\) that passes through the point (1, 1), we can follow these steps:

### Step 1: Rewrite the equation of the circle in standard form
The given equation of the circle is:
\[ x^2 + y^2 - 4x + 4y - 17 = 0 \]
We can rearrange it as:
\[ x^2 - 4x + y^2 + 4y - 17 = 0 \]
To convert it to standard form, we complete the square for the \(x\) and \(y\) terms.

### Step 2: Completing the square
For \(x^2 - 4x\):
\[ x^2 - 4x = (x - 2)^2 - 4 \]
For \(y^2 + 4y\):
\[ y^2 + 4y = (y + 2)^2 - 4 \]
Substituting these back into the equation gives:
\[ (x - 2)^2 - 4 + (y + 2)^2 - 4 - 17 = 0 \]
This simplifies to:
\[ (x - 2)^2 + (y + 2)^2 - 25 = 0 \]
Thus, the standard form of the circle is:
\[ (x - 2)^2 + (y + 2)^2 = 25 \]
This indicates that the center of the circle is at (2, -2) and the radius is 5.

### Step 3: Find the normal line equation
The normal line to the circle at a point \((x_1, y_1)\) can be expressed using the formula:
\[ \frac{x - x_1}{x_1 + g} = \frac{y - y_1}{y_1 + f} \]
where \(g\) and \(f\) are derived from the standard form of the circle. Here, \(g = -2\) and \(f = 2\).

### Step 4: Substitute the point (1, 1) into the normal line equation
We substitute \(x_1 = 1\) and \(y_1 = 1\) into the normal line equation:
\[ \frac{x - 1}{1 - 2} = \frac{y - 1}{1 + 2} \]
This simplifies to:
\[ \frac{x - 1}{-1} = \frac{y - 1}{3} \]
Cross-multiplying gives:
\[ 3(x - 1) = -(y - 1) \]
Expanding this results in:
\[ 3x - 3 = -y + 1 \]
Rearranging leads to:
\[ 3x + y - 4 = 0 \]

### Final Answer
The equation of the normal to the circle that passes through the point (1, 1) is:
\[ 3x + y - 4 = 0 \]