Show that the points $(9,1), (7,9), (-2,12), (6,10)$ are concyclic.

Question

Show that the points $(9,1), (7,9), (-2,12), (6,10)$ are concyclic.

How can we prove that the given points are con-cyclic?

I know the fact the points are said to be concyclic if they lie on the same circle. I substituted the coordinates in the equation of circle and got $4$ equations:-

• $(9-h)^2 + (1-k)^2 = r^2$
• $ (7-h)^2 + (9-k)^2 = r^2$
• $(-2-h)^2 + (-12-k)^2 = r^2$
• $(6-h)^2 + (10-k)^2 = r^2$

Now, here picking the first three equations, I got the centre of circle as $(-8,1)$ and radius = $17$units.

I'm getting no idea what to do further. Is there any short method to solve the question? Please help me here.

Answer 1

Here's a completely different approach. We regard your 4 points as complex numbers $$z_1 = 9 + i,\ \ z_2 = 7 + 9i,\ \ z_3 = -2 + 12i, \ \ z_4 = 6 + 10i$$

and now we calculate the cross ratio of these points. It is a basic fact about the cross ratio that the cross ratio is real if and only if the points are either colinear or concyclic. They are clearly not colinear, just by looking at them, so if the cross ratio $$C(z_1, z_2, z_3, z_4) = \frac{(z_3-z_1)(z_4-z_2)}{(z_3 - z_2)(z_4-z_1)}$$

is real, then we're done. Plugging everything in, we have:$$C(z_1, z_2, z_3, z_4) = \frac{((-2+12i)-(9+i))((6+10i)-(7+9i))}{((-2+12i) -(7+9i))((6+10i)-(9+i))}=\frac{(-11+11i)(-1+i)}{(-9+3i)(-3+9i)}$$

which we can simplify down to

$$=\frac{11}{3}\frac{-2i}{-10i}$$ which is evidently real, since the i's cancel.

Answer 2

The perpendicular bisector of $(9,1)$ and $(7,9)$ is the line $\{(8t+8,2t+5)\mid t\in\Bbb R\}$ and the perpendicular bisector of $(7,9)$ and $(-2,12)$ is the line $\left\{\left(3t+\frac52,9t+\frac{21}2\right)\,\middle|\,t\in\Bbb R\right\}$. They intersect at $C=(0,3)$, which is then the center of the circle passing through $(9,1)$, $(7,9)$ and $(-2,12)$. The distance from $(0,3)$ to $(6,10)$ is $\sqrt{85}$, which the distance from $C$ to the other three points. So, the four points are concyclic. See the picture below.

Answer 3

$\;\;\;\;$ Isosceles trapezoids are cyclic.

Answer 4

Here is a determinant-based formulation.

Start with the fact that a circle through three points $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ has the formula

$$\begin{vmatrix}x^2+y^2 & x & y & 1 \\ x_1^2+y_1^2 & x_1 & y_1 & 1 \\x_2^2+y_1^2 & x_2 & y_2 & 1 \\x_3^2+y_3^2 & x_3 & y_3 & 1 \\ \end{vmatrix}=0$$

Thus if a fourth point $(x_4,y_4)$ lies on the circle it will satisfy

$$\begin{vmatrix}x_4^2+y_4^2 & x_4 & y_4 & 1 \\ x_1^2+y_1^2 & x_1 & y_1 & 1 \\x_2^2+y_1^2 & x_2 & y_2 & 1 \\x_3^2+y_3^2 & x_3 & y_3 & 1 \\ \end{vmatrix}=0$$

or, using a standard row operation to put the points in a more familiar order:

$$\begin{vmatrix}x_1^2+y_1^2 & x_1 & y_1 & 1 \\ x_2^1+y_2^2 & x_2 & y_2 & 1 \\x_3^2+y_3^2 & x_3 & y_3 & 1 \\x_4^2+y_4^2 & x_4 & y_4 & 1 \\ \end{vmatrix}=0$$

This form is a bit unwieldy because it requires evaluating a $4×4$ determinant. We can reduce the problem to a pair of $3×3$ determinants by expanding with minors of the last column. This gives

$$\begin{vmatrix}x_1^2+y_1^2 & x_1 & y_1 \\ x_2^1+y_2^2 & x_2 & y_2 \\x_3^2+y_3^2 & x_3 & y_3 \\ \end{vmatrix}+\color{blue}{\begin{vmatrix}x_1^2+y_1^2 & x_1 & y_1 \\ x_3^1+y_3^2 & x_3 & y_3 \\x_4^2+y_4^2 & x_4 & y_4 \\ \end{vmatrix}}=\begin{vmatrix}x_2^2+y_2^2 & x_2 & y_2 \\ x_3^1+y_3^2 & x_3 & y_3 \\x_4^2+y_4^2 & x_4 & y_4 \\ \end{vmatrix}+\color{brown}{\begin{vmatrix}x_1^2+y_1^2 & x_1 & y_1 \\ x_2^1+y_2^2 & x_2 & y_2 \\x_4^2+y_4^2 & x_4 & y_4 \\ \end{vmatrix}}$$

In the blue matrix interchange the last two rows, which changes the sign of that determinant, and in the brown matrix interchange for first two rows. Then note that the two matrices on the left have two like rows and likewise for the two matrices on the right, so we may combine the unlike rows in each case. We are left with an equality if two $3×3$ determinants, which are much easier to render than the $4×4$ determinant:

$$\begin{vmatrix}x_1^2+y_1^2 & x_1 & y_1 \\ (x_2^1+y_2^2)-(x_4^2+y_4^2) & x_2-x_4 & y_2-y_4 \\x_3^2+y_3^2 & x_3 & y_3 \\ \end{vmatrix}=\begin{vmatrix}x_2^2+y_2^2 & x_2 & y_2 \\ (x_3^1+y_3^2)-(x_1^2+y_1^2) & x_3-x_1 & y_3-y_1 \\x_4^2+y_4^2 & x_4 & y_4 \\ \end{vmatrix}$$

Put in the coordinates of the four points $(9,1),(7,9),(-2,12),(6,10)$:

$$\begin{vmatrix}82 & 9 & 1 \\ -6 & 1 & -1 \\148 & -2 & 12 \\ \end{vmatrix}=82×1×12+(-6)×(-2)×1+148×9×(-1)-82×(-1)×(-2)-(-6)×12×9-(-6)×12-148×1×1=0$$

$$\begin{vmatrix}130 & 7 & 9 \\ 66 & -11 & 11 \\136 & 6 & 10 \\ \end{vmatrix}=130×(-11)×10+66×6×9+136×7×11-130×11×6-66×10×7-136×9×(-11)=0$$

The $3×3$ determinants are equal, so the points are concyclic.

Comment: The common value of the determinants came out zero here, but it may be nonzero for other sets of concyclic points. For instance, the points $(5,0),(0,5),(-5,0),(3,4)$ give a common value of $+250$ rather than zero. The zero value in this problem may be related to two sides of the quadrilateral being parallel.

Answer 5

You can use Ptolemy's inequality to check if the points are concyclic.

Define $A=(9,1),B=(7,9),C=(-2,12),$ and $D=(6,10)$. The inequality states that the points $A$, $B$, $C$, and $D$ are concyclic if $AB \cdot CD + AD \cdot BC = AC + BD$.

By solving for the distances, we have $AB=2\sqrt{17},CD=2\sqrt{17}, AC=11\sqrt{2}, BD=\sqrt{2}, AD=3\sqrt{10}$, and $BC=3\sqrt{10}$. You can check that the inequality becomes an equality, which means they are concyclic.

Answer 6

For 4 distinct points $A,B,C,D,$ if $\,\cos ABC= \frac {(A-B)\cdot (C-B)}{\|A-B\|\cdot \|C-B\|}=-\frac {(A-D)\cdot (C-D)}{\|A-D\|\cdot \|C-D\|}=-\cos ADC$ then the angles $ABC+ADC=\pi,$ which is not possible unless $A,B,C,D$ are concyclic.