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There are three circles, all of them touching each other. The bottom two circles are laying on an imaginary floor, such that they touch the line g=-r as well.

Given are all three radii, r1 (A), r2 (B) and r3 (C). Assuming circle A has its center on (0/0), B has its center on (2 sqrt(r1 * r2), r2 - r1). I am now supposed to find the coordinates of C.

Is this a know problem and has an easy/straighforward solution? I can't seem to find a nice approach.

enter image description here

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Since you know the radii $r_1$, $r_2$, and $r_3$ and the centers $A$ and $B$, the center $C(x_c,y_c)$ must be a distance $r_1+r_3$ from $A$ and a distance $r_2+r_3$ from $B$, which gives the two equations $$x_c^2+y_c^2=(r_1+r_3)^2$$ and $$(x_c-2\sqrt{r_1r_2})^2+(y_c-(r_2-r_1))^2=(r_2+r_3)^2$$ Solving the system for $(x_c,y_c)$ should give the two possible coordinate pairs for $C$ (one as you've pictured, the other "below" the first two circles).

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  • $\begingroup$ Thanks a lot! I was fiddling around with the law of cosine by using the triangle i picted. Accepted and upvoted. $\endgroup$ Commented Mar 18, 2011 at 15:08
  • $\begingroup$ @fjdumont: I'm sure it can be done with the Law of Cosines—since you know the three side lengths of the triangle shown, you can get the measure of the angle at $A$; knowing where $B$ is, you can get the angle between $\overline{AB}$ and the $x$-axis; using those two angles and the length $AC$, you can get the coordinates of $C$. $\endgroup$ Commented Mar 18, 2011 at 15:12
  • $\begingroup$ This is actually pretty much what I did, but I looked for a nicer solution :P $\endgroup$ Commented Mar 18, 2011 at 18:24

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