I have a geometry problem wherein a square contains a quarter-circle A (square sidelength = A's radius) and circle B is inscribed in the square s.t. it's tangent to the bottom/right walls & A. I only know A's radius.
I can successfully solve for B's area, using the sqrt(2) rule shown in the picture.
However, I want to add another circle, C.
C is in the bottom-left of the square, tangent to the outside of B, tangent to the inside of A, and tangent to the bottom wall. How can I find C's area/radius?
I can't see how to find C's radius, or even the angle at which it's tangent to B & A. I'm sure there's some general solution that could find an infinite series of circles (following this pattern) but I can't see the relation.
[EDIT]: I've discovered this is a special case ("Type 9") of Apollonius' problem, and the solution is probably there. I'll close this if I find a solution!
Best Answer
Fix Cartesian coordinates by placing the origin of circle A at $(0,0)$. Center of circle B is at $(-r_B,r_B)$ (using the obvious symmetry that it's along the secondary diagonal at direction -45 degrees on the compass). Requirement of tangency: there is a point $(-t,t)$ that is on the rim of circle B and also on the rim of circle A. The Euclidean distance to the center of A requires $$ (-t)^2+t^2=a^2 \rightarrow t=a/\surd 2. $$ The Euclidean distance of the common tangent point to the center of B requires $$ (-t-(-r_B))^2+(t-r_B)^2=r_B^2. $$ $$ 2t^2-4tr_B+2r_B^2=r_B^2. $$ Inserting $t$ from above $$ r_B^2-2\sqrt{2}ar_B+a^2=0 \rightarrow r_B=a(\sqrt{2}\pm 1). $$ Discard the upper sign because we require $r_B<a$, so $r_B=(\sqrt{2}-1)a$. Center of circle C is at coordinates $(-x_C,r_C)$, $x_C>0$. Using the radius $r_C$ as the y-coordinate ensures that circle C is tangent to the horizontal. The circle C is rolling inside the circle A at a distance $r_C$, so it has distance $a-r_C$ to the center. Define the compass angle $\gamma$, $\pi/2 < \gamma < \pi$, in circular coordinates such that the center of C is at $$ \left( \begin{array}{c} -x_C \\r_C \end{array} \right) = (a-r_C) \left( \begin{array}{c} \cos\gamma\\ \sin\gamma \end{array} \right) . $$ Distance between centers of C and B must be $r_C+r_B$, according to Pythagoras $$ [-x_C-(-r_B)]^2+[r_C-r_B]^2=(r_C+r_B)^2; $$ $$ [(a-r_C)\cos\gamma - (-r_B)]^2 + [r_C - r_B]^2 = (r_C+r_B)^2. $$ $$ (a-r_C)^2\cos^2\gamma +2r_B(a-r_C)\cos\gamma +r_B(r_B-4r_C)=0. $$ Eliminate $\gamma$ via the second of the equations above, $r_C=(a-r_C)\sin\gamma$, $\cos\gamma = -\sqrt{1-\sin^2\gamma}<0$, $$ (a-r_C)^2[1-(\frac{r_C}{a-r_C})^2] -2r_B(a-r_C)\sqrt{1-(\frac{r_C}{a-r_C})^2} +r_B(r_B-4r_C)=0 $$ $$ (a-r_C)^2-r_C^2 -2r_B\sqrt{(a-r_C)^2-r_C^2} +r_B(r_B-4r_C)=0 $$ $$ 2r_B\sqrt{(a-r_C)^2-r_C^2} =a^2-2ar_C+r_B^2-4r_Cr_B $$ $$ 4r_B^2(a^2-2ar_C) =(a^2-2ar_C+r_B^2-4r_Cr_B)^2 ; $$ and this is a quadratic equation for $r_C$ which can be solved numerically, inserting $r_B=a(\sqrt{2}-1)$, $$ 4(a+2r_B)^2 r_C^2 +4(-a^3-2a^2r_B-4r_B^3+ar_B^2)r_C +(r_b-a)^2(r_B+a)^2=0. $$ $$ 4a^2(9-4\sqrt{2}) r_C^2 +8a^3(9-7\sqrt{2})r_C +4a^4(3-2\sqrt{2})=0. $$
$$ r_C/a= \frac{\sqrt{2}-1}{9-4\sqrt{2}} =\frac{5\sqrt{2}-1}{49} \approx 0.123899$$ The angle $\gamma$ is then computed via $\gamma=\arcsin\frac{r_C}{a-r_C}=\arcsin\frac{\surd 2}{10}\approx 171.86989^\circ$