`x^(2)+y^(2)-6x+8y=56`
What is the area of a circle whose equation is given above?

Options

A. `25pi`
B. `81pi`
C. `162pi`
D. `6,561pi`

Answer

To find the area of the circle given by the equation \(x^2 + y^2 - 6x + 8y = 56\), we first need to rewrite the equation in the standard form of a circle. The standard form of a circle's equation is:

\[ (x - h)^2 + (y - k)^2 = r^2 \]

where \((h, k)\) is the center of the circle and \(r\) is the radius.

### Step 1: Rearranging the equation

We start with the given equation:

\[ x^2 + y^2 - 6x + 8y = 56 \]

We can rearrange it to:

\[ x^2 - 6x + y^2 + 8y - 56 = 0 \]

### Step 2: Completing the square for \(x\)

For the \(x\) terms \(x^2 - 6x\), we complete the square:

1. Take half of the coefficient of \(x\) (which is \(-6\)), square it: \((-6/2)^2 = 9\).
2. Add and subtract this square inside the equation:

\[ x^2 - 6x + 9 - 9 \]

This gives us:

\[ (x - 3)^2 - 9 \]

### Step 3: Completing 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: \((8/2)^2 = 16\).
2. Add and subtract this square inside the equation:

\[ y^2 + 8y + 16 - 16 \]

This gives us:

\[ (y + 4)^2 - 16 \]

### Step 4: Substitute back into the equation

Now, substituting back into the equation, we have:

\[ (x - 3)^2 - 9 + (y + 4)^2 - 16 - 56 = 0 \]

Combine the constants:

\[ (x - 3)^2 + (y + 4)^2 - 81 = 0 \]

This simplifies to:

\[ (x - 3)^2 + (y + 4)^2 = 81 \]

### Step 5: Identify the radius

From the standard form \((x - h)^2 + (y - k)^2 = r^2\), we can see that:

- The center of the circle is \((3, -4)\).
- The radius \(r\) is \(\sqrt{81} = 9\).

### Step 6: Calculate the area of the circle

The area \(A\) of a circle is given by the formula:

\[ A = \pi r^2 \]

Substituting the radius:

\[ A = \pi (9)^2 = 81\pi \]

### Final Answer

Thus, the area of the circle is:

\[ \boxed{81\pi} \]