Find the center of a third externally tangent circle

Question

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.

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. — amd, Feb 27 '18 at 21:35
You might find this link helpful. https://math.stackexchange.com/questions/256100/how-can-i-find-the-points-at-which-two-circles-intersect — Narlin, Mar 01 '18 at 22:53

John Glenn · Answer 1 · 2018-02-27T22:33:40.197

1

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:

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

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

edited Feb 27 '18 at 22:33

answered Feb 27 '18 at 21:44

John Glenn

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So this works only when $P_{1y} = P_{2y}$? That would not be general enough for my use case. – user1562293 Feb 27 '18 at 22:12
Yes, but it is only a matter of incorporating the possibility of $P_{1y}\neq P_{2y}$ – John Glenn Feb 27 '18 at 22:16
But can you verify if the answer indeed holds for $P_{1y}=P_{2y}$? – John Glenn Feb 27 '18 at 22:18
In the yellow box I think $P_3$ should be $P_1$ – user1562293 Feb 27 '18 at 22:30
Noted. It's been changed – John Glenn Feb 27 '18 at 22:35
So what is $\theta$ equal to? – user1562293 Feb 27 '18 at 22:36
Well, $\theta$ largely depends on what the values for the radii of the first two circles, since $A,B$ and $C$ will vary according to the lengths of the sides of triangles. – John Glenn Feb 27 '18 at 22:40
All the radii are known. The only unknown variable is $P3$. Is it not possible to solve for $P3$ so that only the three radii and $P1$, $P2$ would appear in the solution? – user1562293 Feb 27 '18 at 22:42
If in fact all values needed are known, then you can use the law of sines to find $A$, taking some time to find $\beta$, you'll get $\theta=A-\beta$ – John Glenn Feb 27 '18 at 22:58

Find the center of a third externally tangent circle

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