From a point A(1, 1) on the circle `x^(2)+y^(2)-4x-4y+6=0` two equal chords AB and AC of length 2 units are drawn. The equation of chord BC, is

Options

A. 4x+3y-12=0
B. x+y=4
C. 3x+4y=4
D. x+y=6

Explanation

To solve the problem step-by-step, we will follow these steps:

### Step 1: Rewrite the Circle Equation
The given equation of the circle is:
\[ x^2 + y^2 - 4x - 4y + 6 = 0 \]
We can rewrite it in standard form by completing the square.

### Step 2: Complete the Square
1. Group the x and y terms:
\[ (x^2 - 4x) + (y^2 - 4y) + 6 = 0 \]
2. Complete the square for x:
\[ x^2 - 4x = (x - 2)^2 - 4 \]
3. Complete the square for y:
\[ y^2 - 4y = (y - 2)^2 - 4 \]
4. Substitute back into the equation:
\[ (x - 2)^2 - 4 + (y - 2)^2 - 4 + 6 = 0 \]
\[ (x - 2)^2 + (y - 2)^2 - 2 = 0 \]
\[ (x - 2)^2 + (y - 2)^2 = 2 \]

### Step 3: Identify the Center and Radius
From the standard form, we can identify:
- Center \( C(2, 2) \)
- Radius \( r = \sqrt{2} \)

### Step 4: Verify Point A
The point \( A(1, 1) \) lies on the circle, as substituting \( x = 1 \) and \( y = 1 \) gives:
\[ (1 - 2)^2 + (1 - 2)^2 = 1 + 1 = 2 \]

### Step 5: Find the Length of Chords AB and AC
Given that the lengths of chords \( AB \) and \( AC \) are both 2 units, we can use the property of chords in a circle.

### Step 6: Use the Right Triangle Property
The two equal chords \( AB \) and \( AC \) create a right triangle \( APQ \) with \( A \) as the vertex. The distance from point \( A \) to the center \( C \) is:
\[ AC = \sqrt{(2 - 1)^2 + (2 - 1)^2} = \sqrt{1 + 1} = \sqrt{2} \]

### Step 7: Calculate the Distance from A to Chord BC
Using the Pythagorean theorem, we can find the distance \( d \) from point \( A \) to the chord \( BC \):
\[ AP^2 + d^2 = AC^2 \]
Where \( AP = 1 \) (half the length of the chord):
\[ 1^2 + d^2 = (\sqrt{2})^2 \]
\[ 1 + d^2 = 2 \]
\[ d^2 = 1 \]
\[ d = 1 \]

### Step 8: Find the Slope of Chord BC
Since \( A(1, 1) \) and \( C(2, 2) \) are symmetric with respect to the line \( y = x \), the slope of chord \( BC \) will be the negative reciprocal of the slope of line \( AC \):
- The slope of line \( AC \) is \( 1 \), so the slope of line \( BC \) is \( -1 \).

### Step 9: Write the Equation of Chord BC
Using point-slope form:
\[ y - 2 = -1(x - 2) \]
Simplifying:
\[ y - 2 = -x + 2 \]
\[ x + y = 4 \]

### Conclusion
The equation of chord \( BC \) is:
\[ x + y = 4 \]