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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.

Michael Rozenberg
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geo_freak
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    It sounds like you just got one of the bisector equations wrong: could it be $y = \frac12 x - 3$ instead of $+3$? Also, even if you solve $x=4$ and $x=2y-6$ you should not get $y=12$, but $y=5$. – Erick Wong Sep 14 '17 at 23:11
  • @ErickWong Thank you catching two mistakes. Looking back at it, yes, I did erroneously but $-3$ instead of $+3$, and $y = 5$, just as you said. – geo_freak Sep 14 '17 at 23:13
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    Some tips on tracing the error: since you already have the correct answer $(4,-1)$, you can first check to see whether it lies on the two bisector lines you found. If it doesn't lie on one of the bisectors then that bisector must be wrong. Also, you can test whether a particular bisector is correct by picking a random point on it and seeing if the distances from that point to the two points (the ones it's supposed to bisect) are equal. – Erick Wong Sep 14 '17 at 23:14
  • Thank you so much for helping me catch those mistakes! It helped me answer the question correctly, and I will definitely try to use the tips that you have provided $\ddot\smile$. – geo_freak Sep 14 '17 at 23:16
  • One way to tell, is to view them as complex numbers, i.e. $9+4i$ instead of $(9,4),$ and then find their cross-ratio. If they are either concyclic or collinear, then the cross ratio is real. Or else it is $\infty.$ (It can be $\infty$ in either the concyclic case or the colinear case.) If they are neither concyclic nor collinear, then the imaginary part of the cross-ratio is not $0. \qquad$ – Michael Hardy Sep 14 '17 at 23:28
  • . . . . . and another way is the approach you're already taking. – Michael Hardy Sep 14 '17 at 23:30
  • Does this answer your question? Can't find an equation for calculating when 4 moving points have a circle passing through them – Clemens Bartholdy Aug 22 '21 at 12:08

5 Answers5

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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:

enter image description here

lhf
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    very good point! still the observation about the sign is not accurate enough, as you may still permut a,b,c. i guess you want them oriented in a certainway, so that the proper coefficient is positive. – orangeskid Sep 14 '17 at 23:57
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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.

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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.

Hari Shankar
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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

SimonVazquez
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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.

Michael Rozenberg
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