Find the area of the circle whose equation is given as \(x^{2} + y^{2} - 4x + 8y + 11 = 0\).
Answer Details
To find the area of the circle whose equation is given as \(x^{2} + y^{2} - 4x + 8y + 11 = 0\), we need to first complete the square for both \(x\) and \(y\).
Rearranging the equation, we get:
\begin{aligned} x^2 - 4x + y^2 + 8y &= -11 \\ x^2 - 4x + 4 + y^2 + 8y + 16 &= 9 \\ (x-2)^2 + (y+4)^2 &= 3^2 \end{aligned}
Now we can see that the equation represents a circle with center \((2, -4)\) and radius \(3\).
The formula for the area of a circle is \(A = \pi r^2\), where \(r\) is the radius. In this case, the radius is \(3\), so the area is:
\begin{aligned} A &= \pi r^2 \\ &= \pi (3)^2 \\ &= \boxed{9\pi} \end{aligned}
Therefore, the answer is \(\boxed{9\pi}\).