How can I prove that just one circle passes through three points?. I, at some way, asked it in a forum, but no one answered me, so I thought could make a forum. Thanks in advance.
[Math] how many circles can pass through three points? (demonstration)
circlesgeometryproof-writingtriangles
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THIS ANSWER GIVES YOU THE FAMILY OF CIRCLES THROUGH TWO GIVEN POINTS.
The way I would work is something like this... if the two points $P_1(x_1,y_1)$ and $P_2(x_2,y_2)$ are on your circle then the line segment $[P_1P_2]$ is a chord. Suppose that the distance $|P_1P_2|=2d$.
Now the perpendicular from the centre of a circle to a chord bisects the chord. We can get the equation of this perpendicular bisector because we have that the midpoint of $[P_1P_2]$ is on it... $((x_1+x_2)/2,(y_1+y_2)/2)$. The slope of the perpendicular bisector, $m$, satisfies $$m\cdot m_{[P_1P_2]}=-1,$$ so is
$$m=-\frac{x_2-x_1}{y_2-y_1},$$
so the perpendicular bisector has equation
$$y=-\frac{x_2-x_1}{y_2-y_1}x+\frac{x_2-x_1}{y_2-y_1}\frac{x_1+x_2}{2}+\frac{y_1+y_2}{2},$$
briefly $k=mh+c$ where $k\sim y$, $h\sim x$ and
$$m=-\frac{x_2-x_1}{y_2-y_1},$$
and
$$c=\frac{x_2-x_1}{y_2-y_1}\frac{x_1+x_2}{2}+\frac{y_1+y_2}{2}$$
Here $h$ is free --- each $h$ gives you a different centre $(h,k)=(h,mh+c)$.
By drawing a picture you will see a RAT triangle with vertices at, say the midpoint of $[P_1P_2]$, the centre $(h,mh+c)$ and $P_1$. Using Pythagoras we have that
$$r^2=d^2+\left(\frac{x_1+x_2}{2}-h\right)^2+\left(\frac{y_1+y_2}{2}-(mh+c)\right)^2$$
So we have that the circles in question are:
$$\{(x-h)^2+(x-(mh+c))^2=r^2:h\in\mathbb{R}\},$$
where each of $m$ and $c$ are functions of $P_1,\,P_2$ and --- once $h$ is chosen --- $r$ is a function of $P_1$, $P_2$ and $h$.
The operation that takes a point $P$ on the circle and find the other intersection point of the line $(AP)$ with the circle is a real automorphism of the circle.
The composition of $3$ such automorphisms is still a real automorphism. If you identify the circle with $\Bbb P^1(\Bbb R)$, it corresponds to a real homography $t \mapsto \frac {at+b}{ct+d}$.
In theory, given $3$ points on the circle and their images you can determine the parameters $a/d,b/d,c/d,\ldots \in \Bbb P^1(\Bbb R)$ and from then construct the line going through the two fixpoints (the parameters of the line are rational fractions in $a,b,c,d$ so they're a rational fraction in the coordinates of the three image points).
There probably even is a way to get a construction that doesn't depend on which $3$ points you choose on the circle, but at the moment I haven't checked on how to construct that line.
Given a point $A \in \Bbb P^2(\Bbb R)$, call $\sigma_A$ the involution of the circle obtained by taking a point $M$ and intersecting the line $(MA)$ with the circle.
Say an automorphism of the circle is direct if the output points turn in the same direction as the input point and indirect if it's not the case (this given by the sign of the determinant $ad-bc$).
The $\sigma_A$ are indirect involutions so they don't represent all the indirect automorphisms (a space of dimension $3$ while we have only $2$ dimensions by picking $A$). However, any direct automorphism can be given in infinitely many ways as a composition of two $\sigma_{A_i}$.
The fixpoints (possibly over $\Bbb C$) of $\sigma_A \circ \sigma_B$ are clearly the (possibly complex) intersections of $(AB)$ with the circle, so for any $C \in (AB)$ there is a unique point $D$ such that $\sigma_A \circ \sigma_B = \sigma_C \circ \sigma_D$. Constructing $D$ or $C$ from the other one is really easy, for example $C$ is the intersection of $(AB)$ with the line $(\sigma_D(M))( \sigma_A \circ \sigma_B (M))$, for any $M$ on the circle.
Next, we clearly have the simplification $\sigma_A \circ \sigma_A = id$.
So given $4$ points we can simplify $(\sigma_A \circ \sigma_B) \circ (\sigma_C \circ \sigma_D)$ by looking at the intersection of $(AB)$ and $(CD)$ (which is possibly at infinity) and moving both $B$ and $C$ to that point while changing $A$ and $D$ appropriately to $A'$ and $D'$. This gives us $(\sigma_A \circ \sigma_B) \circ (\sigma_C \circ \sigma_D) = \sigma_{A'} \circ \sigma_{D'}$
So now that we know how to represent direct automorphisms of the circle with couples of points and compute compositions, and because there is a direct relation between the fixpoints and the line formed by the points, all we have to do is to compute $\sigma_C \circ \sigma_B \circ \sigma_A \circ \sigma_C \circ \sigma_B \circ \sigma_A$ and simplify it to some $\sigma_X \circ \sigma_Y$. Then $(XY)$ intersects the circle at the two fixpoints $D_1$ and $D_2$ of $\sigma_C \circ \sigma_B \circ \sigma_A$.
Note that this construction should work if you replace the circle with any nondegenerate conic, so ellipses, parabolas, and hyperbolas too. You can also generalize this to an odd number $n$ of points instead of $3$, though the complexity of the construction increases linearly with $n$.
Best Answer
Say you have point $A,B,C$. Consider the fact that the center of the circle $O$ has the same distance from $A,B,C$, then it has to lie on the vertical bisector of both $AB$ and $BC$. Assuming $A,B,C$ not on a same line, there 's only one $O$ possible.