Three Circles Meeting at One Point

Question

We have three triples of points on the plane, that is, $X=\{x_1, x_2, x_3\}$, $Y=\{y_1, y_2, y_3 \}$, and $Z=\{z_1, z_2, z_3\}$, where $x_i, y_i, z_i$ are points on the plane. I was wondering if there is a simple (or not-so-simple) algebraic relation satisfied by the coordinates of the 9 points if we know that three circles determined by $X, Y, Z$ pass through a single point.

Answer 1

Here is one approach to obtain a polynomial condition.

1. For a point $(x,y)\in\mathbb R^2$ you can compute a “circle vector” as follows: $$(x,y)\mapsto(1+x^2+y^2, 1-x^2-y^2, 2x, 2y)$$ Do this for all nine points.
2. Write down the resulting vectors for set $X$ as the rows of a $3\times 4$ matrix. Then compute all $3\times3$ determinants you can obtain by omiting one of the columns. So the first element of the resulting vector would be the determinant of all columns except the first and so on. This results in a single four-element vector which encodes the circle through $X$. Do the same for $Y$ and $Z$.
3. Write down these three vectors, one from each set, as the rows of a matrix, and do the same thing over again. This gives you a single four-element vector.
4. Square all coordinates of that vector, then negate the first and sum them up. If the result is zero, your circles intersect in a single point.

Here is why this works. This approach makes use of Möbius circle geometry. There you can encode a circle with center $(x,y)$ and radius $r$ as a coordinate vector

$$\begin{pmatrix}1+(x^2+y^2-r^2)\\1-(x^2+y^2-r^2)\\2x\\2y\end{pmatrix}$$

or any multiple thereof (so these are homogeneous coordinates). Two circles $a=(a_1,a_2,a_3,a_4)$ and $b=(b_1,b_2,b_3,b_4)$ intersect one another orthogonally iff $$-a_1b_1+a_2b_2+a_3b_3+a_4b_4=0$$ You can verify this by doing the computation; you'll find that the equation is equivalent to $(x_a-x_b)^2+(y_a+y_b)^2=r_a^2+r_b^2$. A point is a circle of radius zero. It is orthogonal to itself as well as to any “real” circle on which it lies. So the first step is turning the points into vectors of this form, with $r=0$. Then we are looking for a vector to encode the corresponding circle. The determinants give the right magnitudes but “wrong” sign patterns. In particular, the sign of the third coordinate will not match all the other signs. But these “errors” will cancel out when you do this determinant thingy one more time, so the final result is the circle orthogonal to the three specified circles, encoded according to the rules above. It will be a point if it is orthogonal to itself, which is what you compute by adding squared coordinates, negating the first.

Answer 2

Simple - probably not.

Not-so-simple...

By solving a quadratic equation, you can determine the two points of intersection of the circles $X$ and $Y$. Then the Ptolemy's theorem gives the condition that one of those 2 points lies on the circle $Z$.

PS. Suppose the 3 circles intersect at $O$. An inversion centered in $O$ turns the three circles into three straight lines, so we get 3 collinearity conditions, 2 of which determine the center of inversion and the 3rd is the condition above. This might provide an alternative route for the original problem you are trying to solve (assuming it is different from the question you asked).

Answer 3

Let $\triangle ABC$ have edge-lengths $a=|BC|$, $b=|CA|$, $c=|AB|$. Suppose circles of radius $u$, $v$, $w$ about respective points $A$, $B$, $C$ meet at a common point $P$.

Coordinatizing, we can write $A=(0,0)$, $B=(c,0)$, $C=(b\cos A, b\sin A)$, $P=(p,q)$, so that $$\begin{align} u^2 &= p^2 + q^2 &(1)\\ v^2 &= (p-c)^2 + q^2 = p^2 + q^2 - 2 p c + c^2 &(2)\\ w^2 &= (p-b \cos A)^2 + (q-b\sin A)^2= p^2 + q^2-2b p \cos A - 2 b q \sin A + b^2 &(3) \end{align}$$

Use $(1)$ to eliminate $p^2+q^2$ from $(2)$ and $(3)$:

$$\begin{align} u^2 - v^2 + c^2 &= 2 p c &(4)\\ u^2 - w^2 + b^2 &= 2 b p \cos A + 2 b q \sin A &(5) \end{align}$$

Then, solve the system $(4)$, $(5)$ for $p$ and $q$. Substituting this solution into $(1)$, replacing $\sin^2 A$ with $1-\cos^2 A$, and then writing $\cos A = \frac{1}{2bc}(-a^2+b^2+c^2)$, gives this polynomial condition for the concurrence of the three circles: $$\begin{align} 0 &= a^2 b^2 c^2 + a^2 u^4 + b^2 v^4 + c^2 w^4 \\ &\quad- \left(\;a^2 u^2 + v^2 w^2\;\right)\left(-a^2 + b^2 + c^2\;\right) \\ &\quad- \left(\;b^2 v^2 + w^2 u^2\;\right)\left(\phantom{-}a^2 - b^2 + c^2\;\right) \\ &\quad- \left(\;c^2 w^2 + u^2 v^2\;\right)\left(\phantom{-}a^2 + b^2 - c^2\;\right) &(\star)\end{align}$$

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To answer your question, you "only" need to determine centers $A$, $B$, $C$, distances, $a$, $b$, $c$, and radii $u$, $v$, $w$ from your point-sets $X$, $Y$, $Z$, and then make appropriate substitutions into $(\star)$. Doing this symbolically is straightforward, if tedious ---see, for instance, the MathWorld "Circumcircle" article--- and the result will be an enormous polynomial. As a practical matter, it's probably best to compute the various elements numerically, and then substitute.