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