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
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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
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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
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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
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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?