Technique for proving four given points to be concyclic?

Question

While making my way through an exercise, I stalled on question 7:

7. Prove that the points $(9, 6)$, $(4, -4)$, $(1, -2)$, $(0, 0)$ are concyclic.

The book does not provide any guidance on how to tackle such a question and I can only assume that the authors are assuming that anybody using their textbook will have covered this in a previous course.

I have looked around online for resources that could help me, but none of the ones I've found seem to offer a thorough explanation, building from the ground up.

Would anybody here happen to know where I can find a good explanation of how to prove such a thing?

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

let us call the points $$A = (0,0), B= (1,-2), C = (4, -4), D = (9, 6)$$ you need to vrify that $$2\cos \angle BCD = \frac{BC^2 + CD^2 - BD^2}{BC \cdot CD} = -\frac{AB^2 + AD^2-BD^2}{AB \cdot AD} = -2\cos \angle BAD $$

Answer 5

Yet another way is to evaluate the determinant $\begin{vmatrix} |A|^2 & A_x & A_y & 1 \\ |B|^2 & B_x & B_y & 1 \\ |C|^2 & C_x & C_y & 1 \\ |D|^2 & D_x & D_y & 1 \end{vmatrix}$. It's zero iff the matrix annihilates some nonzero vector $\begin{pmatrix} k\\l\\m\\n \end{pmatrix}$, and it annihilates that vector iff $k|P|^2 + lP_x + mP_y + n = 0$ for all $P \in \{A, B, C, D\}$, and that equation defines a circle (or a line or point, which are degenerate circles).

This is less elegant than some other solutions in that it requires Cartesian coordinates. On the other hand, it's manifestly symmetric in its treatment of the four points.

Answer 6

From this answer here, the condition is given as:

$$ \det\begin{bmatrix} 9^2+6^2 & 9 & 6 & 1 \\ 4^2+(-4)^2 & 4 & -4 & 1 \\ 1+2^2 & 1 & -2 & 1 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} = 0 $$

Answer 7

Let $P_i$, $i=1,...4$ be your points. Let $P'_k$, $k=2,3,4$ be the inversions of the $P_k$ with respect to a unit circle centered on $P_1$. The original four are concyclic if the $P'_k$ are collinear. This will be true when the determinant of the following 3x3 is zero:

Det[{{ $x'_2$, $x'_3$, $x'_4$ } , { $y'_2$, $y'_3$, $y'_4$ } , { 1, 1, 1 }}] = 0?

Note this method applies to any collection of N points. Invert N-1 with respect to a unit circle centered on the first one, and check if the N-1 inversions are collinear.