Technique for proving four given points to be concyclic?

You can use Ptolemy's theorem:

A quadrilateral is inscribable in a circle if and only if the product of the measures of its diagonals is equal to the sum of the products of the measures of the pairs of opposite sides.

In our case, it is obvious from mental diagram that diagonals are $\overline{(9, 6)(1, -2)}$ and $\overline{(0, 0)(4, -4)}$, and

$$|\overline{(9, 6)(1, -2)}|\cdot |\overline{(0, 0)(4, -4)}|=8\sqrt2\cdot4\sqrt2=\color{#c00}{64}$$

and

$$|\overline{(9, 6)(0, 0)}|\cdot |\overline{(1, -2)(4, -4)}|+|\overline{(0, 0)(1, -2)}|\cdot |\overline{(9, 6)(4, -4)}|=$$ $$=\sqrt{117}\cdot\sqrt{13}+\sqrt5\cdot\sqrt{125}=\sqrt{9\cdot13}\cdot\sqrt{13}+\sqrt5\cdot\sqrt{5\cdot25}=39 + 25 = \color{#c00}{64}$$

so points are concyclical.

It is enough to find two opposite vertices whose angles add to 180 degrees. Make vectors of the sides, and use the dot product to calculate cosines of the vertex angles. The cosines of opposite vertices need to be equal in magnitude, but opposite in sign.

Use the property that the perpendicular bisectors of two cords on a circle intersect at the centre. A line passing through $(9,6)$ and $(4,-4)$ is $2x-12$. The perpendicular bisector of that segment is thus $-\frac12x+{\frac {17}{4}}$. Likewise, the line passing through $(0,0)$ and $(4,-4)$ is $-x$, and its perpendicular bisector is $x-4$. The intersection of those two lines is at the point $\left(\frac{11}2,\frac32\right)$. From there just test that the distance to all the points from the centre are equal and you'll find they are concyclic.