Calculate the area enclosed by the curve x^2 + y^2 - 10x + 4y - 196 = 0.

The coefficients of $x^2$ and $y^2$ are equal, therefore, the curve $x^2 + y^2 - 10x + 4y - 196 = 0$ is a circle.
 

$x^2 + y^2 - 10x + 4y - 196 = 0$

$(x^2 - 10x) + (y^2 + 4y) = 196$

$(x^2 - 10x + \color{#FC6255}{25}) + (y^2 + 4y + \color{#FC6255}{4}) = 196 + \color{#FC6255}{25} + \color{#FC6255}{4}$

$(x - 5)^2 + (y + 2)^2 = 225$

$(x - 5)^2 + (y + 2)^2 = 15^2$
 

The standard equation of circle with center at (h, k) and radius r is...
$(x - h)^2 + (y - k)^2 = r^2$

 

Thus, for the given circle, r² = 225
$A = \pi r^2 = 225 \pi ~ \text{unit}^2$   ←   answer