The equation of the tangent to the circle `x^(2)+y^(2)-4x+4y-2=0` at (1,1) is

Options

A. `x-3y+2=0`
B. `x+3y-2=0`
C. `3x+y-1=0`
D. `x+3y=4`

Answer

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

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

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

We can rearrange it as:

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

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

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:

\[ (x - 2)^2 - 4 + (y + 2)^2 - 4 - 2 = 0 \]

This simplifies to:

\[ (x - 2)^2 + (y + 2)^2 - 10 = 0 \]

Thus, we can write the equation of the circle as:

\[ (x - 2)^2 + (y + 2)^2 = 10 \]

### Step 3: Identify the Center and Radius
From the standard form, we can identify the center of the circle \((h, k) = (2, -2)\) and the radius \(r = \sqrt{10}\).

### Step 4: Use the Tangent Line Formula
The formula for the tangent line to a circle at the point \((x_1, y_1)\) is given by:

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

Where \(g\), \(f\), and \(c\) are coefficients from the circle's equation. From our equation, we have:

- \(g = -4\)
- \(f = 4\)
- \(c = -2\)

Substituting \(x_1 = 1\) and \(y_1 = 1\):

\[ x(1) + y(1) + (-4)(1 + 1) + 4(1 + 1) - 2 = 0 \]

This simplifies to:

\[ x + y - 8 + 8 - 2 = 0 \]

### Step 5: Simplify the Equation
Now simplifying gives:

\[ x + y - 2 = 0 \]

### Step 6: Rearranging the Equation
Rearranging gives:

\[ x - 3y + 2 = 0 \]

### Final Answer
Thus, the equation of the tangent to the circle at the point (1, 1) is:

\[ \boxed{x - 3y + 2 = 0} \]