Showing That Points Are Concyclic The question is as follows: 

Show that the points (-1, 4), (9, 4), (3, -8), and (11,0) are concyclic. 

I know that the intersection of the perpendicular bisectors of at least two chords will be the center of the circle. I found the perpendicular bisector of the points $(-1, 4)$ and $(9, 4)$ and the perpendicular bisector of the points $(9, 4)$ and $(11, 0)$. The equation for the perpendicular bisector of the first two points were $x = 4$, however, I am not sure whether that is true or not, and for the second two points, I got the equation for the perpendicular bisector to be $y = \frac{1}{2}x + 3$, and then I solved this equation in terms of $x$, and got $x = 2y-6$. When I set $x = 4$ and $x =2y - 6$ equal to each other, I got the $y$ value to be 12. But that is incorrect because the correct answer is $(4, -1)$. 
Also, I even drew the two lines and their perpendicular bisectors and they both intersected at $(4, -1)$. 
I am pretty sure that there may be something with my algebra, but I just can't trace the error. Any help will be greatly appreciated. 
 A: Another approach is to use incircle test:

Four points $a,b,c,d$ are cocircular iff the determinant below is zero:
  $$
\begin{vmatrix}
x_a&y_a&x_a^2+y_a^2&1\\
x_b&y_b&x_b^2+y_b^2&1\\
x_c&y_c&x_c^2+y_c^2&1\\
x_d&y_d&x_d^2+y_d^2&1\\
\end{vmatrix}
$$

The sign of the determinant tells us whether $d$ is inside, on, or outside the (oriented) circle determined by $a,b,c$.
Here is a picture illustrating this, from the classic paper Primitives for the Manipulation of General Subdivisions and the Computation of Voronoi Diagrams by Guibas and Stolfi:

A: For showing four points are concyclic, one can use complex numbers. If $z_0, z_1, z_2, z_3$ are complex numbers, the points represented by them are concyclic if and only if 
$$\frac{(z_0-z_1)(z_1-z_3)}{(z_0-z_2)(z_0-z_3)}$$ is real. In this case, taking $z_0 = -1+4i, z_1 = 9+4i, z_2 = 3-8i, z_4 = 11$, the above ratio is 
$$\frac{(-4+12i)(-2+4i)}{(6+12i)(-12+4i)} = \frac{1}{4}$$
and hence the points are concyclic.
A: We need to prove that $$AB\cdot DC+AD\cdot BC=AC\cdot BD$$ or
$$10\cdot\sqrt{128}+\sqrt{160}\cdot\sqrt{20}=\sqrt{160}\cdot\sqrt{180},$$
which is obvious.
A: Contributing the potpourri of approaches here: Labelling the points A,B,C, D in the order given by you, its easy to find the intersection of $AB,CD$ as $P(15,4)$
Now $PA \times PB = 96 = PC \times PD$. Hence $ABDC$ is a cyclic quadrilateral.
The method suggested by Muralidharan may be the most economical in general.
A: Calling (xi,yi) to the four points and (x0,y0) the unknown center. We write the 4 eq of the circle and substract two at two having just 2 eq 2 unkowns because the squared vanish at substraction and also R vanishes, then
2(x2-x1)  xo + 2(y2-y1)  y0  =  x2^2 - x1^2 + y2^2 - y1^2
2(x4-x3)  xo + 2(y4-y3)  y0  =  x4^2 - x3^2 + y4^2 - y3^2
the system incompatible if they were not in a circle
