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Given two tangent circles $C_1=(P_1, r_1)$ and $C_2=(P_2, r_2)$ find the centerpoint $P_3$ (its coordinates) of a third circle with a known radius $r_3$ that is externally tangent to both $C_1$ and $C_2$.

I can easily solve this on paper by creating two new circles in $P_1$ and $P_2$ with radii $r_1+r_3$ and $r_2+r_3$ respectively and finding their intersections.

I am however unable to find the coordinates $P_{3x}$ and $P_{3y}$ analytically. I have arrived at a system of quadratic equations but I was unable to solve them and I thought there must be an easier way.

There is a diagram on Wolfram that shows something similar but I could not figure out how to use it to my advantage.

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    $\begingroup$ There is no royal road to enlightenment. This system of equations is quite easy to solve. Notice that the coefficients of the squared terms are all 1. If you subtract the equation of one circle from another, you get the equation of the line on which their intersections lie, so the system will reduce to computing intersections of lines and circles. $\endgroup$ – amd Feb 27 '18 at 21:35
  • $\begingroup$ You might find this link helpful. math.stackexchange.com/questions/256100/… $\endgroup$ – Narlin Mar 1 '18 at 22:53
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General formula: Suppose that $CQ\bot AQ$, where $Q$ is not on $AB$. Therefore, $AQ$ splits $\angle CAB$ such that: $A=\theta+\beta$

By the law of sines, we know that: $$\frac{a'+b'}{\sin C}=\frac{a'+c'}{\sin B}$$ $$ AC=a'+c'=(a'+b')\cdot\frac{\sin B}{\sin C}$$

It follows that: $$\sin \theta=\frac{CQ}{(a'+b')\cdot\frac{\sin B}{\sin C}}$$ $$CQ=(a'+b')\cdot\frac{\sin \theta\sin B}{\sin C}$$ $$CQ=AC\cdot\sin \theta \tag{1}$$ Now, logically, $AQ$ must be equal to below: $$AQ=\sqrt{AC^2-CQ^2}$$ $$AQ=\sqrt{\Biggl((a'+b')\cdot\frac{\sin B}{\sin C}\Biggr)^2-\Biggl((a'+b')\cdot\frac{\sin \theta\sin B}{\sin C}\Biggr)^2}$$ $$AQ=(a'+b')\cdot\frac{\sin B}{\sin C}\cdot\sqrt{1- \sin ^2 \theta}$$ $$AQ=AC\cdot\sqrt{1-\sin ^2 \theta} \tag{2}$$

Recalling $(1)$ & $(2)$, then $(P_{3x},P_{3y})$ should be equal to:

$$(P_{3x},P_{3y}) \Rightarrow (P_{1x}+AQ,P_{1y}+CQ)$$

$$\therefore (P_{1x}+AC\cdot\sqrt{1-\sin ^2 \theta},P_{1y}+AC\cdot\sin \theta)$$

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  • $\begingroup$ So this works only when $P_{1y} = P_{2y}$? That would not be general enough for my use case. $\endgroup$ – user1562293 Feb 27 '18 at 22:12
  • $\begingroup$ Yes, but it is only a matter of incorporating the possibility of $P_{1y}\neq P_{2y}$ $\endgroup$ – John Glenn Feb 27 '18 at 22:16
  • $\begingroup$ But can you verify if the answer indeed holds for $P_{1y}=P_{2y}$? $\endgroup$ – John Glenn Feb 27 '18 at 22:18
  • $\begingroup$ In the yellow box I think $P_3$ should be $P_1$ $\endgroup$ – user1562293 Feb 27 '18 at 22:30
  • $\begingroup$ Noted. It's been changed $\endgroup$ – John Glenn Feb 27 '18 at 22:35

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