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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?

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    $\begingroup$ Does concyclic just mean there's a circle containing those points? If so, have you learned that any three non-colinear points determine a circle? If so, one naive method would be to find that circle for three of the given points, and check that the fourth is on it. $\endgroup$
    – pjs36
    Apr 26, 2015 at 17:29
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    $\begingroup$ Um, I don't actually know for sure, this question just appears in the exercise. I assume it means that all four of them lie on the same circle - if that's what you're saying? A bit like 'collinear' $\endgroup$
    – Au101
    Apr 26, 2015 at 18:26

9 Answers 9

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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.

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    $\begingroup$ Well that is not completly true. Quadrilateral must also be convex. So you have to check that also. $\endgroup$
    – nonuser
    Dec 30, 2021 at 18:05
  • $\begingroup$ @nonuser Can a concave quadrilateral ever be concyclic? $\endgroup$ Jun 21, 2023 at 19:56
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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.

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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.

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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.

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  • $\begingroup$ This is called the InCirce test. It's used in Delaunay triangulations. $\endgroup$
    – lhf
    Sep 1, 2015 at 17:22
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let us call the points $$A = (0,0), B= (1,-2), C = (4, -4), D = (9, 6)$$ you need to vrify that $$2\cos \angle BCD = \frac{BC^2 + CD^2 - BD^2}{BC \cdot CD} = -\frac{AB^2 + AD^2-BD^2}{AB \cdot AD} = -2\cos \angle BAD $$

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    $\begingroup$ Okey dokey, that sounds straightforward enough to do; but could you explain why we do that? $\endgroup$
    – Au101
    Apr 26, 2015 at 20:31
  • $\begingroup$ @Au101, i am using the cosine rule and the fact that the opposite angles add up to $180^\circ.$ when the aangles are supplementary, their cosines are equal. $\endgroup$
    – abel
    Apr 26, 2015 at 20:34
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    $\begingroup$ Maybe I'm dense, but why are opposite angles adding up to 180 degrees? And for that matter, what are 'opposite' angles in this context? We don't know that any two of these points are forming a diameter... $\endgroup$ Apr 26, 2015 at 20:37
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    $\begingroup$ @StevenStadnicki, for a quadrilateral to be cyclic, it is necessary and sufficient for the opposite angles to be supplementary. the diameter of the quad dont have to be the diameters of the circumcircle. the angle condition follows from the fact that angle subtended by an arc on the circumference is half that subtended at the center. $\endgroup$
    – abel
    Apr 26, 2015 at 20:40
  • $\begingroup$ Hi. I tried to edit, but couldn't do it. I have left it the way I found it. Do you mind editing it? $\endgroup$
    – aarbee
    Dec 9, 2020 at 6:45
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To prove such questions just do few simple steps:--

  • Find two equations of opposite lines

  • Let's say:

  • $a_1x+b_1y+c_1=0$ and

  • $a_2x+b_2y+c_2=0$

  • Now what you have to check is whether

  • $a_1•a_2$ = $b_1•b_2$ or not

  • If yes, points are con-cyclic and not then they aren't con-cyclic

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    $\begingroup$ Hi. Do you mind giving a link (reference) to this property? $\endgroup$
    – aarbee
    Dec 9, 2020 at 6:49
  • $\begingroup$ @Lol Olo That does seem to work, but is there any proof for that? $\endgroup$ Mar 1, 2021 at 18:24
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Find the equation of the circumference going through the points $(4,-4), (1,-2)$ and $(0,0)$. Given that the general form of the circumference is $x^2+y^2+ax+by+c=0$, the following conditions must be met:

  1. $4^2+(-4)^2+4a-4b+c=0$
  2. $1^2+(-2)^2+a-2b+c=0$
  3. $0^2+0^2+0a+0b+c=0$

From which we infer that $a=-11$ and $b=-3$. Hence the required equation is $x^2+y^2-11x-3y=0$. Now we have to verify that this circumference cointains point $(9,6)$ as well. Plugging the values into the equation we get this: $9^2+6^2-11\cdot 9 -3\cdot 6=81+36-99-18=0$. Therefore, the four points all belong to the same circumference whose equation is $x^2+y^2-11x-3y=0$.

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From this answer here, the condition is given as:

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

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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.

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