euclidean-geometrygeometrymetric-spacesplane-geometry
Given are two intersecting circles $A, B$ with radius $r_A > r_B$ and center $M_A, M_B$. A third circle $C$ with radius $r_C$ and center $M_C$ fits in the non intersecting part of circle $B$ and touches the two circles $A, B$ at the points $P_{CA}, P_{CB}$. What are the results for $M_C, P_{CA}, P_{CB}$ for given $A, B, r_C$?
Not part of question, only thougts on solving:
Does the path for $M_C$ describe a part of an ellipse? This could be the first step to the solution.
By scaling one can assume that $R_a=1$, and we can put $R_b=r$. This answer only looks at the case where the second circle of radius $r$ is positioned so that its point of tangency with the unit circle (center at $A=(0,0)$ lies in the first quadrant. We also assume that the $r$ circle protudes enough in the $x$ direction so that the bbox needs to extend to the right of $x=1$, and that the $r$ circle is far enough down from the top that the entire $r$ circle lies below the line $y=1$.
The center of the $r$ circle lies at $((1+r) \cos \theta, (1+r) \sin \theta)$ where $\theta$ is the angle formed by the center of the $r$ circle, the origin, and the positive $x$ axis. Then the condition that the $r$ circle lies below $y=1$ is $$[1]\ \ \ r+(1+r)\sin \theta \le 1.$$ And the condition that the $r$ circle protrudes beyond the line $x=1$ is $$[2]\ \ \ r+(1+r)\cos \theta \ge 1.$$ The ratio $k$ under these assumptions comes from the fact that, in this situation, the height of the bbox is $2$ while its width, extending from $x=-1$ to the rightmost point of the $r$ circle, is $1+(1+r)\cos \theta +r$. This gives the ratio as $$k = \frac{2}{(1+r)(1+\cos \theta)},$$ which solves for $\cos \theta$ as $$\cos \theta=\frac{2}{k(1+r)}-1.$$ One can then find $\theta$ via arccos. Even in this simple case, it is necessary to know that [1] and [2] are satisfied, in order for the expression for $\cos \theta$ to achieve $k$ be valid. I've attempted to eliminate $\cos \theta$ and $\sin \theta$ from conditions [1] and [2], but it gets involved.
Note that here the value of $k$ must lie in the interval $[1/(1+r),1]$ in order for there to be a possible $r$ circle, where the smallest $k$ corresponds to the case where the $r$ circle is tangent to the unit circle at $(1,0)$. One naturally also needs $r<1$ or else the $r$ circle will extend beyond $y=1$ and change the formula for $k$.
It remains to look at other cases, for example the case wherein the $r$ circle extends both to the right of $x=1$ and above $y=1$. And the conditions [1] and [2] need to be restated without $\theta$, for a complete answer. There may be another way to look at this question, which would not involve so many cases and conditions.
You can find a formula for the area of intersection between two circles here. Using only this formula, you can find the area you are looking for.
In the figure above I've drawn the scenario you describe, with names for the various areas which make up the intersection area between the circles. If we denote the Area of Intersection between the two circles with radii $r_1$ and $r_2$ as $AI(r_1,r_2)$, we see that: $$A_2 + 2 \cdot A_1 = AI(r_c,r_a) - AI(r_d, r_a) = C_1$$ $$A_3 + 2 \cdot A_1 = AI(r_a,r_c) - AI(r_b, r_c) = C_2$$ $$A_2 + A_3 + 2 \cdot A_1 = AI(r_c,r_a) - AI(r_d, r_b) = C_3$$ Therefore, $$C_1 + C_2 - C_3 = 2 \cdot A_1$$ or $$A_1 = \frac{C_1 + C_2 - C_3}{2}$$
Best Answer
I have put a Geogebra sheet at https://www.geogebra.org/m/qnFkgnev. I am going to write the 4 equations needed to solve for the smallest circle. A convention that I am using: If I have fixed the value of a point, then it is addressed as $(x(P),y(P))$, whereas, if I have the point as an unknown, it is addressed as $(P_x,P_y)$ In the Geogebra sheet, you can move point C about but the browser is slow. So, click, hold down and move cursor to a new spot on the circle, release. Wait a few seconds and hopefully all will recalculate/redraw. My 4 unknowns are $D_x,D_y,E_x,E_y$
Unknown point D, is on the line between B & C. m is the slope of that line. $$m=\left(\frac{y(B)-y(C)}{x(B)-x(C)}\right)$$ The equation: $$D_y=y(C)+m(D_x-x(C))$$
The radius of the small circle is the same between C & D and between E & D. $$|C-D|=|E-D|$$
Point E has to lie on circle A. $$E_x^2+E_y^2-3E_x-3E_y=4.5$$ (I think this circle is hard coded in the geogebra sheet, hence the numerical values shown in this equation.)
Point D must lie on a line from A through E. $$m_1=\left(\frac{E_y-y(A)}{E_x-x(A)}\right)$$ and the equation: $$D_y=E_y+m_1(D_x-E_x)$$
Once these were solved (by Geogebra CAS), I could move the fixed point C about and a new point D was generated. I saved these using $SetValue(D_n,D)$. Next I chose 3 of the D values and put a circle through them. Surprise (or not), that circle does not quite go through all of the D values. i.e. the locus of points in question do not quite make the arc of a circle. Is it an ellipse? Maybe, but that certainly isn't proven yet.