What is the area enclosed by the circle $x^{2} + y^{2} - 4x - 12 = 0$?

Explanation

1. 01

   Complete the square: $(x - 2)^{2} + y^{2} = 16$, a circle of radius 4 with centre $(2, 0)$.

2. 02

   Moving a region does not change its area: $2\int_{-2}^{6} \sqrt{16 - (x - 2)^{2}}\,dx = 16\pi$ square units.