# A Find a circle inside of and tangent to a larger circle

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1. Oct 5, 2017

### votingmachine

I don't think that is quite the proof I was looking for. I'm not sure if I was clear. The question I had was how do we know the line from $\dot b$ to $\dot p$ ends at the tangent point $\dot g$, and not another point on the large circle. If it is shown they are the same point, then the similar triangles completes the radius derivation.

I don't disbelieve they are the same point. The drawings line up and it FITS. Every drawing (and the helpful MP4 graphic) LOOKS right. I just don't quite see why.

2. Oct 5, 2017

### andrewkirk

@votingmachine Start with the diagram with $\dot c,\dot p, \vec d$ marked, and let $\dot f,\dot g$ be the centre and tangent point of the circle that is the solution. Now define $\dot b$ to be the point where the line from $\dot g$ through $\dot p$ crosses the circle again.

Then we can easily show that the triangles $\dot g\dot b\dot c$ and $\dot g\dot p\dot f$ are similar (since both are isosceles, and they share an angle), from which it follows that $\dot b\dot c$ is parallel to $\dot p\dot f$, which is parallel to $\vec d$.

So, to find this solution, all we need do is

1. find the point $\dot b$ where the line through $\dot c$ parallel to $\vec d$ but in reverse direction cuts the circle
2. find the point $\dot g$ where the line from $\dot b$ through \dot p$cuts the circle again. That is the tangent point 3. find the point$\dot f$where the radius from$\dot g$to$\dot c$cuts the ray in direction$\vec d$from$\dot p##. That is the centre of the smaller circle.
which is the approach Ventrella described - albeit in slightly different words..

3. Oct 5, 2017

### votingmachine

Got it. That method explicitly draws triangles, and they are similar. Thanks.

4. Oct 6, 2017

### Baluncore

B–K1 is a diameter of the big circle. Thales' theorem gives the angle at G between B and K1 as 90°.
P–K2 is a diameter of the small circle. Thales' theorem gives the angle at G between P and K2 as 90°.
Since the diameters B–K1 and P–K2 are parallel, and the angles at G are the same, the point K2 must lie on the small circle and they must be similar triangles.