Find the area of the circle whose equation is given as \(x^{2} + y^{2} - 4x + 8y + 11 = 0\).
Equation of a circle: \((x - a)^{2} + (y - b)^{2} = r^{2}\)
Given that \(x^{2} + y^{2} - 4x + 8y + 11 = 0\)
Expanding the equation of a circle, we have: \(x^{2} - 2ax + a^{2} + y^{2} - 2by + b^{2} = r^{2}\)
Comparing this expansion with the given equation, we have
\(2a = 4 \implies a = 2\)
\(-2b = 8 \implies b = -4\)
\(r^{2} - a^{2} - b^{2} = -11 \implies r^{2} = -11 + 2^{2} + 4^{2} =9\)
\(r = 3\)
\(Area = \pi r^{2} = \pi \times 3^{2}\)
= \(9\pi\)
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