`x ^(2) + 6x + y^(2) -8y =171`
The equation of a circle in the xy-plane is shown above. What is the positive difference between the x-and y-coordinates of the center of the circle ?

Answer

To solve the equation \( x^2 + 6x + y^2 - 8y = 171 \) and find the positive difference between the x- and y-coordinates of the center of the circle, we will follow these steps:

### Step 1: Rearrange the equation
We start with the given equation:
\[ x^2 + 6x + y^2 - 8y = 171 \]
We want to rearrange this equation to complete the square for both \(x\) and \(y\).

### Step 2: Complete the square for \(x\)
To complete the square for the \(x\) terms \(x^2 + 6x\):
1. Take half of the coefficient of \(x\) (which is 6), square it:
\[ \left(\frac{6}{2}\right)^2 = 3^2 = 9 \]
2. Add and subtract this square inside the equation:
\[ x^2 + 6x + 9 - 9 \]
This can be rewritten as:
\[ (x + 3)^2 - 9 \]

### Step 3: Complete the square for \(y\)
Now, for the \(y\) terms \(y^2 - 8y\):
1. Take half of the coefficient of \(y\) (which is -8), square it:
\[ \left(\frac{-8}{2}\right)^2 = (-4)^2 = 16 \]
2. Add and subtract this square:
\[ y^2 - 8y + 16 - 16 \]
This can be rewritten as:
\[ (y - 4)^2 - 16 \]

### Step 4: Substitute back into the equation
Now substitute the completed squares back into the equation:
\[ (x + 3)^2 - 9 + (y - 4)^2 - 16 = 171 \]
Combine the constants:
\[ (x + 3)^2 + (y - 4)^2 - 25 = 171 \]
So we have:
\[ (x + 3)^2 + (y - 4)^2 = 196 \]

### Step 5: Identify the center of the circle
From the standard form of the circle \((x - h)^2 + (y - k)^2 = r^2\), we can identify:
- The center of the circle \((h, k)\) is \((-3, 4)\).

### Step 6: Calculate the positive difference
Now, we need to find the positive difference between the x-coordinate and y-coordinate of the center:
\[ \text{Positive difference} = |k - h| = |4 - (-3)| = |4 + 3| = 7 \]

### Final Answer
The positive difference between the x- and y-coordinates of the center of the circle is:
\[ \boxed{7} \]