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How can i find a circle that is tangent to two circles which have the same center? Specifically i'm looking for a circle that will contain the smaller circle. I know how to find the circle whose diameter is equal to the difference between the two radii.

This picture may help.enter image description here

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  • $\begingroup$ Are you just looking for the size of the circle? Because it seems there are an infinite number of circles that would fit this definition. $\endgroup$ Jan 8, 2015 at 3:18
  • $\begingroup$ You may want to look up Apollonius of Perga and "Apollonius's Problem". $\endgroup$
    – KSmarts
    Jan 8, 2015 at 3:34
  • $\begingroup$ actually i should have mentioned that i need the circle to be tangent to circle 2 at a given point. So this circle would be defined by a known point on circle 2 at which point it is also tangent to circle 2. then i need to find where the tangent point is on circle 1. $\endgroup$
    – Adil
    Jan 8, 2015 at 4:17

2 Answers 2

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enter image description here

If the radius of the small circle is $|\overline{OP}| = p$ and the radius of the big circle is $|\overline{OQ}| = q$, then the diameter of the target circle is $p+q$, so that its radius is $\frac12(p+q)$.

Note that the points of tangency $P$ and $Q$, along with $O$ (the common center of the original circles) and $M$ (the center of the target circle), are collinear.

So, to "find" the target circle given point $P$, simply draw $\overrightarrow{PO}$; the intersection with the other circle is $Q$, and the midpoint of $\overline{PQ}$ is $M$. (Or, starting with $Q$, draw $\overrightarrow{QO}$; etc.) In terms of vectors or coordinates (with $O$ not-necessarily the origin), you can write

$$p\;( Q - O ) = -q\;( P - O )$$ so that $$M = \frac12\left(\; P + Q \;\right) = \frac{1}{2p} \left(\; (p+q) O + (p-q) P \;\right) = \frac{1}{2q}\left(\; (p+q) O + ( q - p ) Q \;\right)$$

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Take a look at www.arcenciel.co.uk/geometry/index.htm They have proofs and C# coding for many types of circle tangentiality

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