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I've been struggling to find a way to resolve the following problem:

Let $C_1$ a circle of center $V$ and of radius $r_1$. Let $A$ and $B$ two points outside of $C_1$, and $L$ a line passing by them. Let $W$ a point on $L$. Let $C_2$ a circle of center $W$ passing through $A$ (radius $r_2$ is the distance between $A$ and $W$).

I am looking for the coordinate of the point $W$ so that $C_2$ intersects $C_1$ in a single point (that is that $C_1$ and $C_2$ are externally tangent).

The L line, C_1 and a variable C_2 depending on point W

What do you think? I tried to visualize a logic behind it using GeoGebra but couldn't figure it out. Right now what I'm doing in my code (this is for generative design) is increasing the distance between $A$ and $W$ by small steps until the distance $VW$ is equal to $r_1 + r_2$, but I'm looking for the elegant, geometrical solution!

Cœur
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Teatoon
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3 Answers3

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Construct on $L$ a point $P$ such that $AP=r_1$. Center $W$ of the tangent circle is equidistant from $P$ and $V$, hence it can be found as the intersection between $L$ and the perpendicular bisector of $PV$.

Note that two choices are possible for $P$, one leading to an internally tangent circle, the other to an externally tangent circle.

Intelligenti pauca
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The tangency condition (whether true or false) is preserved if we change the radii of the two circles by the same amount in the appropriate direction:

  • For external tangency: Shrink $C_1$ to a point and expand $C_2$'s radius by $r_1$.
  • For internal tangency: Shrink $C_1$ to a point and shrink $C_2$'s radius by $r_1$.

Hence in both cases the modified $C_2$ can be easily constructed as follows:

  1. Construct point $P$ on $L$ such that $|AP| = r_1$.
  2. Construct circle with centre on $L$ that passes through $P$ and $V$.

Notice that this yields exactly the same solution as in the previous answer, but this abstracted idea of changing object sizes is really useful for all kinds of problems, including in geometry (e.g. constructing a circle tangent to 3 given circles) and in other fields (even in computer science). So it is worth learning to see things this way.

user21820
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With $p = (x,y)$ we have

$$ \cases{ C_1\to \|p-V\|^2 = r_1^2\\ L\to p = A + \lambda\vec v,\ \ \|\vec v\| = 1\\ } $$

and now we construct

$$ \cases{ C_1' \to \|p - V\|^2 = (r_1+r_2)^2\\ L'\to p = A + r_2\vec v }\Rightarrow \|A-V+r_2\vec v\|^2 = (r_1+r_2)^2 $$

then

$$ r_2 = \frac{\|A-V\|^2-r_1^2}{2((V-A)\cdot\vec v+r_1)} $$

Cesareo
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