Its Area

The circle has center at (1,0) and radius of $\sqrt{6}$ . $f(x)=6x-6$ after reducing the Determinant. This line passes through the center of the circle, therefore we are looking for the area of the semicircle which is $3\pi$ . Approximately $\boxed{9.42}$

Applying $C_1 \to C_1 + C_2$ :

$\left| \begin{matrix} 0 & x-1 & 1 \\ -2 & -x & 2x-{ x }^{ 2 } \\ 4 & 3-x & { 2x }^{ 2 }-4x \end{matrix} \right|$

Applying $R_3 \to R_3 + 2 R_2$ :

$\left| \begin{matrix} 0 & x-1 & 1 \\ -2 & -x & 2x-{ x }^{ 2 } \\ 0 & 3(1-x) & 0 \end{matrix} \right|$

Expanding along $C_1$ , we have $f(x) = 6(x-1)$ . This line passes through the centre of the circle and hence divides the circle into two equal parts. Hence, the area above the line is simply $\frac{\pi \cdot 6}{2} = \boxed{3\pi}$