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Consider three disjoint circles not necessarily of same radii. How do you draw the smallest circle enclosing all these three circles? Where is its centre, and what is its radius?

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  • $\begingroup$ Do you assume that the three circles are fixed? Or can they be moved around? In the latter case, it would be a circle packing problem. $\endgroup$ – Axel Kemper Nov 12 '14 at 13:31
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    $\begingroup$ Yes, the circles are fixed. $\endgroup$ – Abhishek Nov 12 '14 at 14:05
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    $\begingroup$ This is known as Apollonius's problem. The Wikipedia article lists a lot of different solution methods. $\endgroup$ – Rahul Nov 12 '14 at 15:33
  • $\begingroup$ @Rahul: Are you sure? Enclosing is not the same as touching. Imagine 3 unit circles in a row. They can be enclosed by touching only two of them. $\endgroup$ – Axel Kemper Nov 12 '14 at 15:48
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    $\begingroup$ A related post: stackoverflow.com/questions/6976125/… The many downvotes indicate the complexity of the problem. $\endgroup$ – Axel Kemper Nov 12 '14 at 17:18
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If your goal is to "draw" the smallest circle and you don't really care about the numerical values of its center and radii too much, You can construct it in a geometrical manner.

Let's say the three circles are centered at $A$, $B$ and $C$ respectively.

There are two possibilities.

  1. The smallest circle is touching two of the circles. In this case, the center of the smallest circle will be collinear with the center of the two circles it touches.
  2. The smallest circle is touching all three circles.

It is clear how to construct the smallest circle in $1^{st}$ case. For the $2^{nd}$ case,

  • Construct a line passing through $A$ and $B$.
  • Let $G$ be the intersection of line $AB$ with circle $A$ on the opposite side of point $B$.
  • Let $H$ be the intersection of line $AB$ with circle $B$ on the opposite side of point $A$.
  • Let $I$ be the mid-point of $G$ and $H$.
  • Construct a hyperbola passing through $I$ having $A$ and $B$ as foci (the red curve in the picture below).
  • Repeat the same procedure to the two other combination of pairs of circles to obtain three hyperbolas (the red, green and blue curves in picture below).

These 3 hyperbolas will intersect at two points $P$ and $R$. The point $P$ is the one which lies on the branch that contains the mid point $I$. It will be the center of another candidate of smallest circle (the orange circle) you want.

At the end, we will obtain four candidates of the smallest circle and it is easy to check which one is the smallest one.

In any event, in the $2^{nd}$ case, one can compute the coordinates of the center $P$ by first figuring out the equations for the three hyperbolas and then determine their intersection. The algebra will be a mess and I'll let you have the fun (if you really want that).

A tales of three circles

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Not a full solution but a sketch:

Let us assume the three disjoint circles are described by:

$$\begin{aligned} (x - x_1)^2 + (y - y_1)^2 &= r_1^2 \\ (x - x_2)^2 + (y - y_2)^2 &= r_2^2 \\ (x - x_3)^2 + (y - y_3)^2 &= r_3^2 \end{aligned}$$

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The three circles are at fixed locations and disjoint. Therefore, we know:

$$\begin{aligned} (x_2 - x_1)^2 + (y_2 - y_1)^2 &\ge (r_1 + r_2)^2 \\ (x_3 - x_2)^2 + (y_3 - y_2)^2 &= (r_2 + r_3)^2 \\ (x_1 - x_3)^2 + (y_1 - y_3)^2 &= (r_1 + r_3)^2 \end{aligned}$$

Assuming that the center of the outer circle is $\{x_m;y_m\}$, we want to know for the three touch-points where the outer circle is tangential to the enclosed circles. The touch-points must be located on radii of the small circles and of the outer circle. As additional constraints, the tangential slopes must be equal between small circle and outer circle in all three touch-points.
Experimenting with GeoGebra, I found that there seems to be unique solution. But I could not find an algebraical expression for it.

The geometric construction of the center point $\{x_m;y_m\}$:

  • Three lines from $\{x_m;y_m\}$ to the three center points intersect the circles in the three touch-points.
  • $\{x_m;y_m\}$ is also the intersection of the perpendicular bisectors of the three touch-points.

If the outer circle just has to enclose but not necessarily touch all circles, the construction does not help.
This case probably has to be solved numerically using a non-linear optimization procedure. The coordinates $\{x_m;y_m\}$ are varied. I would take the middle between the biggest two circles as starting point.

Note:
If all three circles are located with their center points on the same line and have the same radius, there is no finite enclosing outer circle which touches all three circles.

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  • $\begingroup$ Is it possible to qualitatively describe the centre? Like what relation does it hold with the circles? I was told it is the radical centre of the circles, but I didn't understand why. $\endgroup$ – Abhishek Nov 12 '14 at 13:12

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