Intersections of circles in a flat plane, when all radii are perturbed by a constant amount

Question

Consider a flat Euclidean plane and three distinct circles on it with one common intersection point shared by all three circles. If the three radii are perturbed by a constant, call it $\delta$, under what conditions will there be another (real) intersection point common to all three circles?

Algebraically, here is the formulation. Assume that the quadratic system $$(x-x_1)^2+(y-y_1)^2=r_1^2,$$ $$(x-x_2)^2+(y-y_2)^2=r_2^2,$$ $$(x-x_3)^2+(y-y_3)^2=r_3^2,$$ already has a solution, the point $S=(x_s,y_s)$. Let us augment the system, $$(x-x_1)^2+(y-y_1)^2=(r_1+\delta)^2,$$ $$(x-x_2)^2+(y-y_2)^2=(r_2+\delta)^2,$$ $$(x-x_3)^2+(y-y_3)^2=(r_3+\delta)^2.$$ Then the point $S$ solves this augmented system with $\delta=0$.

Under what conditions, do real solutions with $\delta\neq0$ exist? And do they have a closed-form?

Case 1

This is the trivial case. When the three centers lie on a straight line, the system will have two distinct solutions $S$ and $T$, both with $\delta=0$. The three black dots are the centers of the three circles. $S$ is the intersection of all three of these circles but, simultaneously, $T$ is another distinct intersection. Both intersections occur when $\delta=0$.

Case 2

In this case, shown in Figure 2, no matter what we add or subtract from the radii, the three circles will never intersect again, for $\delta\neq0$. We can "confirm" this by solving the quadratic system numerically. There are only two solutions and they are,

• point $(1,1)$ with $\delta=0$,
• point $(0.261204, 0.261204)$ with $\delta=-1.78361$.

The "problem" here is that the radii are too small, or $\delta$ is too large of a negative number in magnitude, whichever way you want to look at it. The three radii, when $\delta=0$, are $$r_1=\sqrt{2}\approx1.4142,$$ $$r_2=1,$$ $$r_3=1,$$ so that when you add $\delta$ with a large negative value, the circles become imaginary with negative radii.

Case 3

In this case, the two numerical solutions are

• point $(1,1)$ with $\delta=0$,
• point $(2.06558, 1.36233)$ with $\delta=1.06017$.

Figure 3A shows the first solution and Figure 3B shows the second solution. In this case, $\delta>0$ so we simply add it to the radii and the circles are all real.

Case 4

In this case, the two numerical solutions are

• point $(1,1)$ with $\delta=0$,
• point $(0.561976, 0.408912)$ with $\delta=-0.719213$.

$\delta$ is negative in this case but its magnitude is small enough, smaller than all of the radii, which means that when we decrease the radii, the radii are all still positive and the circles are real. The three radii in this case are $$r_1=\sqrt{2}\approx1.4142,$$ $$r_2=\sqrt{5}\approx2.23607,$$ $$r_3=\frac{3\sqrt{269}}{50}\approx0.899389.$$

Questions

• There is a very elegant Euclidean geometry theorem here. What exactly is it?
• How would you go about proving this?
• Under what conditions is the intersection $T$ real? Imaginary?
• What is the closed-form for the solutions, if possible?
• Are there always exactly two solutions, if we allow both real and imaginary circles? This is a quadratic system in three variables.
• Any references where this may already have been investigated?

Answer 1

Disclaimer: This solution is not complete. I assume that the three centers do not lie on a straight line, and I also make an assumption about the solution of a certain homogeneous system of linear equations. Furthermore, I do not have an idea how to interpret the results geometrically.

Expand the equations of the augmented system, multiply with $-\frac{1}{2}$ and rearrange everything: \begin{eqnarray} x_1x+y_1y+r_1\delta - \frac{1}{2}\left(x^2+y^2-\delta^2\right) & = & \frac{1}{2}\left(x_1^2+y_1^2-r_1^2\right) \\ x_2x+y_2y+r_2\delta - \frac{1}{2}\left(x^2+y^2-\delta^2\right) & = & \frac{1}{2}\left(x_2^2+y_2^2-r_2^2\right) \\ x_3x+y_3y+r_3\delta - \frac{1}{2}\left(x^2+y^2-\delta^2\right) & = & \frac{1}{2}\left(x_3^2+y_3^2-r_3^2\right) \end{eqnarray} With $u=-\frac{1}{2}\left(x^2+y^2-\delta^2\right)$ and $b_i=\frac{1}{2}\left(x_i^2+y_i^2-r_i^2\right),$ we get $$ \begin{pmatrix} x_1 & y_1 & 1 & r_1 \\ x_2 & y_2 & 1 & r_2 \\ x_3 & y_3 & 1 & r_3 \end{pmatrix} \begin{pmatrix} x \\ y \\ u \\ \delta \end{pmatrix} = \begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix} $$ We have to look for solutions of this underdetermined system of equations that also fulfill the definition of $u.$ If the three centers of the circles do not lie on a straight line, then the matrix has full rank and there is (up to scalar multiples) only one solution to the system $$ \begin{pmatrix} x_1 & y_1 & 1 & r_1 \\ x_2 & y_2 & 1 & r_2 \\ x_3 & y_3 & 1 & r_3 \end{pmatrix} \begin{pmatrix} x_0 \\ y_0 \\ u_0 \\ \delta_0 \end{pmatrix} =0 $$ We can use, for example $$ x_0 = \begin{vmatrix} y_1 & 1 & r_1 \\ y_2 & 1 & r_2 \\ y_3 & 1 & r_3 \end{vmatrix} \;\;\; y_0 = -\begin{vmatrix} x_1 & 1 & r_1 \\ x_2 & 1 & r_2 \\ x_3 & 1 & r_3 \end{vmatrix} \\ u_0 = \begin{vmatrix} x_1 & y_1 & r_1 \\ x_2 & y_2 & r_2 \\ x_3 & y_3 & r_3 \end{vmatrix} \;\;\; \delta_0 = -\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} $$ Note that $\delta_0\neq 0$ if the three centers of the circles do not lie on a straight line.

We already know one of the solutions of the system of equations, which is $(x_S\;\;y_S\;\;u_S\;\; 0)^T.$ Therefore, we can get all solutions of the linear system of equations as follows: $$ \begin{pmatrix} x \\ y \\ u \\ \delta \end{pmatrix} = \begin{pmatrix} x_0 \\ y_0 \\ u_0 \\ \delta_0 \end{pmatrix} \lambda + \begin{pmatrix} x_S \\ y_S \\ u_S \\ 0 \end{pmatrix} $$ Now we have to ensure that our definition of $u$ is consistent with the candidates we get from this formula. We need $u=-\frac{1}{2}\left(x^2+y^2-\delta^2\right),$ which means $$ u_0\lambda+u_S = -\frac{1}{2}\left((x_0\lambda+x_S)^2+(y_0\lambda+y_S)^2 -(\delta_0\lambda)^2\right) $$ Simplifications: \begin{eqnarray} u_0\lambda - \frac{1}{2}\left(x_S^2+y_S^2\right) & = & -\frac{1}{2}\left((x_0\lambda+x_S)^2+(y_0\lambda+y_S)^2 -(\delta_0\lambda)^2\right) \\ -2u_0\lambda +x_S^2+y_S^2 & = & (x_0\lambda+x_S)^2+(y_0\lambda+y_S)^2-(\delta_0\lambda)^2 \\ -2u_0\lambda +x_S^2+y_S^2 & = & x_0^2\lambda^2+2x_0x_S\lambda+x_S^2 +y_0^2\lambda^2+2y_0y_S\lambda+y_S^2-\delta_0^2\lambda^2 \\ -2u_0\lambda & = & x_0^2\lambda^2+2x_0x_S\lambda +y_0^2\lambda^2+2y_0y_S\lambda-\delta_0^2\lambda^2 \\ 0 & = & \lambda\; \left((x_0^2+y_0^2-\delta_0^2)\lambda + 2x_0x_S+2y_0y_S + 2u_0\right) \end{eqnarray} We are not interested in the solution $\lambda = 0$ because it gives us $S,$ which we already know. Therefore, we are interested in $$ (x_0^2+y_0^2-\delta_0^2)\lambda + 2x_0x_S+2y_0y_S + 2u_0 = 0 $$ Further analysis is needed to clarify if $x_0^2+y_0^2 = \delta_0^2$ can happen. If not, we get exactly one "interesting" value for $\lambda,$ which is $$ \lambda = -2\;\frac{x_0x_S+y_0y_S+u_0}{x_0^2+y_0^2-\delta_0^2} $$ This choice of $\lambda$ gives "real" circles, if $r_i + \lambda\delta_0 > 0$ for $i=1,2,3.$

Summary: Under the assumption that the centers do not lie on a straight line, and under the assumption that the homogeneous system of linear equation (above) does not produce a solution with $x_0^2+y_0^2 = \delta_0^2,$ there are no additional solutions (if $x_0x_S+y_0y_S+u_0=0$) or there is exactly one additional solution (if $x_0x_S+y_0y_S+u_0\neq 0$). There is a closed form of the solution, but it is probably incomprehensible, if you put it all into one single formula.

Note: $x_0x_S+y_0y_S+u_0=0$ happens if $S$ and two of the centers are on a straight line, and both centers are on the same side of $S.$ In this case, the two circles only touch at one single point, and this fact does not change for different values of $\delta.$ If the third circle does not have its center on the same line, it will never intersect the other two in their common boundary point for any $\delta\neq 0.$

Edit

In the meantime, I found out that $x_0^2+y_0^2=\delta_0^2$ can happen. But there are no obvious geometrical properties associated with this condition, at least I did not detect any. The scenarios in which this happens look as generic as any other scenario with randomly chosen circles - no alignment of the centers of the circles, no special ratio of the radii or any other obvious property.

Answer 2

Not sure if I understood correctly. Two poles or foci are needed for real concurrency of circles intersection.

The family of all circles through 2 concurrent points Apollonian Circles shown in red (Wiki) are one such possibility. Also the matter is related to the topic of bipolar coordinates where circles in the Apollonian set are co-axal.

In this complex plane Real (half) segments to the family have constant segment lengths between concurrent points. Imaginary (half) segments to the family have constant tangent lengths from the origin.

Answer 3

There is a theorem called Three Conics Theorem, which states that if three conics pass through two given distinct points, say $P$ and $Q$, then the lines joining the other two intersections of each pair of conics are concurrent. If the points $P$ and $Q$ are taken as the points at infinity, then this theorem reduce to the theorem that states that radical lines of three circles, which appear as common chords in OP’s problem, are concurrent in a point $K$ known as the radical center (see $\mathrm{Fig.\space 1}$). The scenario, which is described by OP in the problem statement, is a special case of this theorem, where the mentioned point of concurrency $K$ lies on all three circle. Note that we exclude the special case, in which the centers of the three circle are collinear, from our discussion, because the corresponding $K$ tends to infinity.

As already mentioned, we are interested only in the case where $K$ lies on the circles. Our aim is to deduce the necessary equations for the case of $K$ lying anywhere in the finite 2D-space and then investigate whether we can obtain an expression for $\delta$ in terms of known quantities when $K$ lies on circles. In other word, we use equations of two of the three common chords to determine the coordinates of $K$ and then substitute them in to the equation of any one of the circle to find out what happen when we equate this polynomial of degree two in $\delta$ to zero. Since we already know one of the roots of the ensuing quadratic equation (i.e., $\delta = 0$), finding the other root should be a piece of cake.

To keep the expressions and equations simple and short, we have to transform the given configuration from its original coordinate frame to a more appropriate one. The new coordinate frame must have its origin located at the center of one of the circle while its $y-\text{axis}$ (or $x-\text{axis}$) must pass through the center of another circle as shown in $\mathrm{Fig.\space 2}$.

Assume that we are given three circles drawn in $XY\text{-coordinate frame}$. Since we are aware of their respective radii and the $x\text{-}$ and $y\text{-coordinates}$ of their centers, we can write down their equations. $$\left(X-X_i\right)^2+\left(Y-Y_i\right)^2=r_i^2,\quad i=1,2,3.$$

The same three circles have the following equations in the new $xy\text{-coordinate frame}$. $$(x-x_1)^2+(y-y_1)^2=r_1^2,$$ $$\small{\text{where}\quad x_1=\dfrac{\left(X_1-X_3\right) \left|Y_2-Y_3\right|+\left(Y_1-Y_3\right) \left|X_2-X_3\right|}{\sqrt{\left(X_2-X_3\right)^2+\left(Y_2-Y_3\right)^2}}\quad\text{and}\quad\enspace}$$ $$\small{y_1=\dfrac{-\left(X_1-X_3\right) \left|X_2-X_3\right|+\left(Y_1-Y_3\right) \left|Y_2-Y_3\right|}{\sqrt{\left(X_2-X_3\right)^2+\left(Y_2-Y_3\right)^2}}}$$

$$x^2+(y-y_2)^2=r_2^2,\quad{\small{\text{where}}}\quad {\small{y_2=\dfrac{-\left(X_2-X_3\right) \left|X_2-X_3\right|+\left(Y_2-Y_3\right) \left|Y_2-Y_3\right|}{\sqrt{\left(X_2-X_3\right)^2+\left(Y_2-Y_3\right)^2}}}} $$

$$x^2+y^2=r_3^2$$

Now, let’s find the $\delta$, which forces the common chords to have their common point of intersection on the circles, equations of which are given below. $$\text{ Circle-1}\qquad(x-x_1)^2+(y-y_1)^2=(r_1+\delta)^2 $$ $$\text{ Circle-2}\qquad x^2+(y-y_2)^2=(r_2+\delta)^2\qquad\quad$$ $$\text{ Circle-3}\qquad x^2+y^2=(r_3+\delta)^2\qquad\qquad\qquad $$

When they are expanded, we get, $$\text{ Circle-1}\qquad x^2+y^2-2x_1x-2y_1y=2c_1+\delta^2+2r_1\delta,$$ $$\text{ Circle-2}\qquad x^2+y^2-2y_2y=2c_2+\delta^2+2r_2\delta, \quad\space\space \text{and}$$ $$\text{ Circle-3}\qquad x^2+y^2=2c_3+\delta^2+2r_3\delta,\qquad\qquad\qquad\enspace $$ where $\quad c_1=\dfrac{r_1^2-x_1^2-y_1^2}{2},\quad c_2=\dfrac{r_2^2-y_2^2}{2},\quad\text{and}\quad c_3=\dfrac{r_3^2}{2}$.

The common chords of the circle pair 1-3 and circle pair 2-3 are given below. $$x_1x+y_1y=\left(c_3-c_1\right)+\left(r_3-r_1\right)\delta\tag{1}$$ $$y_2y=\left(c_3-c_2\right)+\left(r_3-r_2\right)\delta\tag{2}\qquad\quad$$

We can deduce an expression for the $y-\text{coordinate }\space\bar{\bar{y}}$ of the point of intersection between the two common chords using (2). $$\bar{\bar{y}}=\dfrac{\left(c_3-c_2\right)+ \left(r_3-r_2\right)\delta}{y_2}\tag{3}$$

To derive the expression for its $x-\text{coordinate }\space\bar{\bar{x}}$, we substitute (3) in (1). $$\bar{\bar{x}}=\dfrac{\left(\left(c_3-c_1\right)y_2-\left(c_3-c_2\right)y_1\right)+ \left(\left(r_3-r_1\right)y_2-\left(r_3-r_2\right)y_1\right)\delta}{x_1y_2}\tag{4}$$

For brevity, we express $\bar{\bar{x}}$ and $\bar{\bar{y}}$ as, $$\bar{\bar{x}}=a_x+b_x\delta\qquad\text{and}\qquad \bar{\bar{y}}=a_y+b_y\delta,\quad\text{where,}\tag{5}$$ $$a_x=\dfrac{\left(c_3-c_1\right)y_2-\left(c_3-c_2\right)y_1}{x_1y_2}, \qquad\quad b_x= \dfrac{\left(\left(r_3-r_1\right)y_2-\left(r_3-r_2\right)y_1\right)}{x_1y_2},$$ $$a_y=\dfrac{c_3-c_2}{y_2}, \qquad\text{and}\qquad b_y=\dfrac{r_3-r_2}{y_2}.$$

Finally, by substituting values of $\bar{\bar{x}}$ and $\bar{\bar{y}}$ from (5) in the equation of circle-3, following quadratic equation in $\delta$ can be obtained. $$\left(b_x^2+b_y^2-1\right)\delta^2+2\left(a_xb_x+a_yb_y-r_3\right)\delta+\left(a_x^2+a_y^2-r_3^2\right)=0 \tag{6}$$

We know that $\delta_1=0$ is one real root of (6) and this is possible if and only if its last term $\left(a_x^2+a_y^2-r_3^2\right)$ is equal to zero. With that, equation (6) is reduced further to. $$\left(b_x^2+b_y^2-1\right)\delta+2\left(a_xb_x+a_yb_y-r_3\right)=0, \tag{7}$$ which gives us the other root, i.e., $$\delta_2 =-2\left(\dfrac{a_xb_x+a_yb_y-r_3}{ b_x^2+b_y^2-1}\right).$$

Since it is known for sure that $\delta_1$ is real, we can state that $\delta_2$ is also real unconditionally. This also shows that, if there exist one configuration of three intersecting circles, than there is a second configuration as well.

When performing the calculations to determine $\delta_2$, first check whether $a_x^2+a_y^2-r_3^2\approx 0$. If that is not the case, then an error or errors present in your calculations of quantities such as $c$ s, $b$ s, $a$ s, etc.

We think we have answered in our text all but one of the questions OP posed at the end of the problem statement. As far as references are concerned, we were able to find only one (a thesis), which is unfortunately either behind a paywall or for members only. You can also try here.

$\underline{\text{Edit}}:\space$ We add below an example to show how to do the calculation.

Answer 4

The quadratic system we want to solve is

$ (x - x_1)^2 + (y - y_1)^2 = (r_1 + \delta)^2 \hspace{15pt}(1) $

$ (x - x_2)^2 + (y - y_2)^2 = (r_2 + \delta)^2 \hspace{15pt}(2) $

$ (x - x_3)^2 + (y - y_3)^2 = (r_3 + \delta)^2 \hspace{15pt}(3) $

This is a quadratic system of $3$ equations in $3$ unknowns $x, y, \delta$.

It has a special structure that allows for cancellation of the quadratic terms, by subtracting pairs of equations from each other.

Substracting $(2)$ from $(1)$ gives

$ 2 x (x_2 - x_1) + 2 y (y_2 - y_1) + 2 \delta (r_2 - r_1) = K_1 \hspace{15pt}(4) $

where $K_1 = r_1^2 - r_2^2 + x_2^2 + y_2^2 - x_1^2 - y_1^2 $

Similarly, we obtain the second equation by subtracting $(3)$ from $(1)$,

$ 2 x (x_3 - x_1) + 2 y( y_3 - y_1) + 2 \delta (r_3 - r_1) = K_2 \hspace{15pt}(5) $

where $K_2 = r_1^2 - r_3^2 + x_3^2 + y_3^2 - x_1^2 - y_1^2 $

Solving this linear system is relatively easy by a computer program, and gives the solution

$ (x, y, \delta) = X_0 + t X_1 \hspace{15pt}(6) $

where $X_0, X_1$ are known, and $t \in \mathbb{R}$ is yet to be determined.

To determine $t$ plug in $(6)$ into $(1)$, and this will give two solutions for $t$ if the discriminant of the quadratic equation in $t$, is positive. But there can be no solutions, if the discriminant is negative.

Equation $(1)$ can be written in matrix format as

$ (p- p_0)^T Q (p - p_0) = 0 $

where $p = (x, y, \delta) $ and $p_0 = (x_1, y_1, - r_1 ) $

and

$ Q = \operatorname{diag}\{1, 1, -1\} $

Pluggin in $(6)$, we get

$ p = X_0 + t X_1 $

So that

$ ( t X_1 + X_0 - p_0)^T Q (t X_1 + X_0 - p_0) = 0 $

which expands to

$ t^2 \ X_1^T Q X_1 + 2 t \ X_1^T Q (X_0 - p_0) + (X_0 - p_0)^T Q (X_0 - p_0) = 0 \hspace{15pt}(7) $

Solving we have a maximum of two solutions. One solution is the one with $\delta =0$, so there will be at most one new solution.

Using Mathematica or a similar symbolic math package, it should possible to obtain $X_0$ and $X_1$ explicitly (i.e. closed-form) in terms of $x_1, y_1, r_1, x_2, y_2, r_2, x_3, y_3, r_3$. Further, it can be possible to express the discriminant of equation (7), in terms of these variables. But I suspect this is too complicated to be of any use.