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We have two given circles (highlighted green in the illustration below). The center of the first circle is $A=(x_A,y_A)$ and its radius is $r_a$. The center of the second circle is $B=(x_B,y_B)$ and its radius is $r_b$.

How can we calculate the center $C=(x_C,y_C)$ of circles which touch the two given ones (as the highlighted orange circle does)? Possibly there exists two curves on which infinitely many center points of such circles lie:

  • one curve on which center points of "small circles" (like the orange) lie
  • one curve on which center points of "big circles" lie (big circles that encompass the two green cicrles)

Here is what I tried: Draw a straight line $AB$ and then mark two points $A'$ and $B'$ with distance $r_C$ each from the periphery of the two given circles on the straight line.

How can I find a simple formula (or even a implicit curve) for the center $C$ of the desired circle(s)?

enter image description here

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    $\begingroup$ There are infinitely many such $C$ and circles. $\endgroup$
    – user10354138
    Commented Sep 12, 2021 at 7:00
  • $\begingroup$ Ok - this is what I not expected but is great. Thank you for this hint! Can we find then a formula for all these circles' center points $C_k$? Thesse center points should form then a line or a curve? $\endgroup$
    – user736865
    Commented Sep 12, 2021 at 7:02
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    $\begingroup$ The $C_k$ all lies on one branch of a hyperbola with foci $A,B$. $\endgroup$
    – user10354138
    Commented Sep 12, 2021 at 7:04
  • $\begingroup$ Interesting! I was more thinking about anotherone "big cicrle" embedding these two green ones as additional solution. For this big embedding circle - does there exist infinitely solutions as well? $\endgroup$
    – user736865
    Commented Sep 12, 2021 at 7:06
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    $\begingroup$ Still hyperbola with the same foci, since this is $CA\pm r_A=CB\pm r_B$, the choice of each $\pm$ depends on whether we want circle $C$ to touch the circle $A$ (or $B$) internally or externally. All four combinations give you the four branches of two hyperbolae (if $r_A=r_B$ you get the perpendicular bisector of $AB$ plus a hyperbola). $\endgroup$
    – user10354138
    Commented Sep 12, 2021 at 7:37

1 Answer 1

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Given disjoint circles, and unequal radii, the locus of centers comprises two hyperbolas. Begin by intersecting the axis with the both circles. Let it intersect circle $A$ at $A_1$ and $A_2$, and circle $B$ at $B_1$ and $B_2$, as shown here, where $A_1$ and $B_1$ are between the two centers.

enter image description here

Let $K$ be the midpoint of $A_2B_2$, and $L$ the midpoint of $A_1B_1$. Let $P$ be the center of a circle externally tangent to both or internally tangent to both. This relation follows:

$(PA - PB)^2 = (r_a-r_b)^2$

The locus of $P$ is a hyperbola with foci $A$ and $B$. Points $K$ and $L$ both satisfy the condition for $P$, and they lie on the axis, so those are the vertices.

enter image description here

Now start again. Let $M$ be the midpoint of $A_2B_1$, and $N$ the midpoint of $A_1B_2$. Let $Q$ be the center of a circle externally tangent to one of the given circles and internally tangent to the other. This relation follows:

$(QA - QB)^2 = (r_a+r_b)^2$

The locus of $Q$ also is a hyperbola with foci $A$ and $B$. This time the vertices are at $M$ and $N$.

Other cases to investigate would be intersecting circles or congruent circles.

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  • $\begingroup$ Thank you for this great illustration! It would be nice if you may add a figure where the red circle completely inscribes the two green circles. As I understand it, the center of such a circle would lie on the left hyperbolic curve shown by your first figure - right?. Or maybe you can extend your first figure by adding such a big circle. $\endgroup$
    – user736865
    Commented Sep 12, 2021 at 10:32
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    $\begingroup$ Yes, as you said, any point on the left branch of that first hyperbola is the center of a circle tangent to both given circles and enclosing both. The sketches are both quite precisely plotted, so you might confirm that by printing it or by pasting the image into some geometry software. I did not wish to clutter the sketch further. In fact, I do not even know why I bothered with those axis intersections and vertices. One other case you might consider would have one given circle entirely within the other. $\endgroup$
    – Pope
    Commented Sep 12, 2021 at 18:58
  • $\begingroup$ Thank you for confirming! May I ask which tool you used for these nice plots? $\endgroup$
    – user736865
    Commented Sep 12, 2021 at 19:00
  • $\begingroup$ That was done with The Geometer's Sketchpad, still my weapon of choice. $\endgroup$
    – Pope
    Commented Sep 12, 2021 at 19:11
  • $\begingroup$ great - will try this too :-) $\endgroup$
    – user736865
    Commented Sep 12, 2021 at 19:25

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