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)?
Best Answer
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.
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.
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.