I am having some difficulties answering this question:

For some fixed $x_0,x_1,y_0,$ and $y_1$ in $\mathbb R$, where the ordered pairs $(x_0,y_0) \neq (x_1,y_1)$,prove that the following sets have an intersection of precisely one element:

  1. $S_1=\left\{(x,y) \in \mathbb R \times \mathbb R: \sqrt{(x-x_1)^2+(y-y_1)^2}=\sqrt{(x_1-x_0)^2+(y_1-y_0)^2} \right\}$

  2. $S_2=\left\{(x,y) \in \mathbb R \times \mathbb R: \sqrt{\left(x-(2x_1-x_0)\right)^2+\left(y-(2y_1-y_0)\right)^2}=\sqrt{\left(2\cdot(x_1-x_0)\right)^2+\left(2\cdot(y_1-y_0)\right)^2}\right\}$

These sets describe the following picture, where $(x_2,y_2)$ is simply $(2x_1-x_0,2y_1-y_0)$:

Picture of Small Circle Inside a Large Circle Making 1 Point of Contact

The question at hand is equivalently framed as finding the solution to the set of equations:

\begin{align} &(1) \quad \sqrt{(x-x_1)^2+(y-y_1)^2}=\sqrt{(x_1-x_0)^2+(y_1-y_0)^2} \\ &(2)\quad\sqrt{\left(x-(2x_1-x_0)\right)^2+\left(y-(2y_1-y_0)\right)^2}=\sqrt{\left(2\cdot(x_1-x_0)\right)^2+\left(2\cdot(y_1-y_0)\right)^2} \end{align}

If $(x,y)$ satisfies $(1)$ and $(2)$, then we must have $(1')$ and $(2')$:

\begin{align} &(1')\quad(x-x_1)^2+(y-y_1)^2=(x_1-x_0)^2+(y_1-y_0)^2 \\ &(2')\quad\left(x-(2x_1-x_0)\right)^2+\left(y-(2y_1-y_0)\right)^2=\left(2\cdot(x_1-x_0)\right)^2+\left(2\cdot(y_1-y_0)\right)^2 \end{align}

After expanding $(2')$, I was able to find that some portion of the terms on the left side of the equality had the form $(x-x_1)^2+(y-y_1)^2$, so I proceeded to substitute in the right hand side of equation $(1')$. This effectively substitutes a constant in for the quadratic terms $x^2$ and $y^2$, while keeping the $x$ and $y$ terms, which will allow us to implicitly solve for one of the variables. After some additional simplification, I produced the equation:

$$(3) \quad (x-x_1)(x_0-x_1)+(y-y_1)(y_0-y_1)=(x_0-x_1)^2+(y_0-y_1)^2$$

Some additional algebra and factoring leads to the equation that is giving me problems:

$$(4) \quad (x_0-x_1)(x-x_0)+(y_0-y_1)(y-y_0)=0$$

It is easy to see from $(4)$ that $(x_0,y_0)$ is a valid solution (which makes sense from the construction of the sets). However, I am having difficulties showing that $(x_0,y_0)$ is the only valid solution. When we initially assumed that $(x_0,y_0) \neq (x_1,y_1)$, we equivalently have that $x_0 \neq x_1 \text{ OR } y_0\neq y_1$.

Consider the case when $x_0=x_1$ and $y_0 \neq y_1$. From $(4)$, we see that, although we must have $y=y_0$, $x$ can equal any number $\in \mathbb R$. This is a problem if we want to show that only one ordered pair satisfies both equations. A similar complication arises if we assume that $x_0\neq x_1$ and $y_0 = y_1$.

As far as I can tell, all of the algebraic manipulations I made are perfectly reversible. Therefore, if $(x,y)$ satisfies $(4)$, it should also satisfy $(2)$. However, that is clearly not the case. i.e. it is definitely FALSE that $\forall x \in \mathbb R : (x,y_0) \in S_2$.

Where have I gone wrong in the argument?

  • $\begingroup$ You did not finish solving system $(1')-(2')$: you must solve $(4)$ for $y$ and substitute into $(1')$, then solve the result for $x$. $\endgroup$ Mar 20, 2022 at 16:32
  • $\begingroup$ @Intelligentipauca hmmm. Okay - that is what I originally started to do, but it looked like I was going to get a nasty equation. I'll take another look: thank you! $\endgroup$ Mar 20, 2022 at 16:53
  • 1
    $\begingroup$ Otherwise, you could prove that statement in a geometrical way, using triangular inequality. Very fast and no algebra at all. $\endgroup$ Mar 20, 2022 at 18:26
  • $\begingroup$ @Intelligentipauca I've tried for a while to simplify the resulting equation following the substitution. I am having no luck reducing it to an appropriate form. Regardless, I am having some difficulties understanding why $(4)$ has the problems that I mentioned. As far as I can tell, all of the algebraic manipulations I made are perfectly reversible. Therefore, if $(x,y)$ satisfies $(4)$, it should also satisfy $(2)$. However, that is clearly not the case. i.e. it is definitely FALSE that $\forall x \in \mathbb R : (x,y_0) \in S_2$. Am I wrong in my thinking? $\endgroup$ Mar 20, 2022 at 22:06
  • $\begingroup$ You have a SYSTEM of equations: your system $(1)-(2)$ is equivalent to system $(1)-(4)$, or to system $(2)-(4)$. That's all you can tell. $\endgroup$ Mar 20, 2022 at 22:26

2 Answers 2


This can undoubtedly be attacked and solved in a much more elementary and direct way, but:

According to Bézout’s Theorem on the intersection of plane curves, if $C$ and $C'$ are curves in the projective plane $\Bbb P^2(\Bbb C)$ over $\Bbb C$ of degrees $d,d'$ respectively, then there are exactly $dd'$ points of intersection, counting multiplicity.

Now a circle $A(x^2+y^2)+Bx+Cy+D=0$, when homogenized, becomes $A(X^2+Y^2)+BXZ+CYZ+DZ^2=0$, which necessarily has the two points “at infinity” $I=(1,i,0)$ and $J=(1,-i,0)$. Thus, in $\Bbb P^2(\Bbb C)$, every two distinct circles $C,C'$ automatically intersect at $I$ and $J$. When they are tangent (but still distinct), that’s a point of intersection of multiplicity two, and this fills up the count of four. There will be no other intersections.

To continue beyond your question, what are the other possibilities for the intersection of two circles? They may intersect in two distinct points that you see; or they may have different centers but not intersect visibly, in which case their intersection is a pair of conjugate complex points; or they may be concentric. In this last case, the two are tangent at $I$ and $J$.

You can check all this out with pencil-and-paper computations of your own.

  • $\begingroup$ I think this will be fairly inaccessible to me for the foreseeable future, but thank you for the attention nonetheless! $\endgroup$ Mar 20, 2022 at 23:28
  • $\begingroup$ But surely you can dream up examples of pairs of conics with $0,1,2,3,4$ real (finite) intersections but have trouble dreaming up a pair with five intersections. $\endgroup$
    – Lubin
    Mar 21, 2022 at 19:48

Here is an alternative approach using a result taken from here: Analytical proof that the diameter is the longest chord of a circle.

From the linked proof, we have the following result (formatted to describe the posted scenario):

Lemma: For a given $(a_0,b_0)$ on the circumference of a circle $C$ centered at $(a_1,b_1)$, the point $(a_2,b_2)$ on the circumference that maximizes the Euclidean distance to $(a_0,b_0)$ the point that is rotated $180 \deg$ relative to $(a_0,b_0)$. One way of expressing this $(a_2,b_2)$ is $(2a_1-a_0,2b_1-b_0)$. Importantly, the distance between $(a_0,b_0)$ and $(2a_1-a_0,2b_1-b_0)$ is $2$ times as large as the Euclidean distance between $(a_0,b_0)$ and $(a_1,b_1)$. Further, for the given $(a_0,b_0)$ ordered pair, $(2a_1-a_0,2b_1-b_0)$ is the only ordered pair on the circumference of $C$ with this maximum value. The line segment connecting these points necessarily passes through the center $(a_1,b_1)$ and is therefore the diameter.

Here is how we use this information to our advantage:

Using the notation from the original post, we can see that $(x_0,y_0) \in S_1$, which means that $(x_0,y_0)$ lays on the circumference of a circle centered at $(x_1,y_1)$. Next, referencing the system of equations, notice that the constants to the right of the equality sign for the $S_1$ and $S_2$ expressions are scaled versions of one another. Specifically, setting $d_1=\sqrt{(x_1-x_0)^2+(y_1-y_0)^2}$, it is straightforward to show that $\sqrt{\left(2\cdot(x_1-x_0)\right)^2+\left(2\cdot(y_1-y_0)\right)^2}=2d_1$. These results place us in the appropriate context to use our lemma.

Now, apply our above lemma to the point $(x_0,y_0)$ for the circle that is centered on $(x_1,y_1)$. Then we have that $(2x_1-x_0,2y_0-y_1)$ is the only point on the circumference of the circle centered at $(x_1,y_1)$ that is twice the Euclidean distance between $(x_0,y_0)$ and $(x_1,y_1)$. Notice that this implicitly provides us with $(2x_1-x_0,2y_0-y_1) \in S_1$ because $(2x_1-x_0,2y_0-y_1)$ lays on the circumference of the circle centered at $(x_1,y_1)$.

The Euclidean distance between $(x_0,y_0)$ and $(x_1,y_1)$ is $d_1$, which means the Euclidean distance between $(x_0,y_0)$ and $(2x_1-x_0,2y_1-y_0)$ is $2d_1$. Crucially, $(x_0,y_0)$ is the only such point on the circumference of the circle centered at $(x_1,y_1)$ that is $2d_1$ away from $(2x_1-x_0,2y_1-y_0)$. In order to be a member of $S_2$, however, the object must be $2d_1$ away from $(2x_1-x_0,2y_1-y_0)$. This means that $(x_0,y_0)$ is the only member of $S_1$ that is simultaneously in $S_2$.

Thus, we have demonstrated that $S_1 \cap S_2$ contains only one element, and that element is $(x_0,y_0)$.


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