# Three Tangent Circles

Three circles (O_{i}), with centers at O_{i}, _{k} denotes the point of contact of (O_{i}) and (O_{j}). P_{i}P_{j} extended crosses again (O_{i}) in Q_{k} and (O_{j}) in R_{k}. The circle (A_{k}) tangent to (O_{i}) in Q_{k} and (O_{j}) in R_{k} have an interesting property.

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Copyright © 1996-2018 Alexander BogomolnyThree circles (O_{i}), with centers at O_{i}, _{k} denotes the point of contact of (O_{i}) and (O_{j}). P_{i}P_{j} extended crosses again (O_{i}) in Q_{k} and (O_{j}) in R_{k}. Let r_{i} denote the radius of circle (O_{i}).

The center A_{k} of circle (A_{k}) lies at the intersection of lines C_{i}Q_{k} and C_{j}R_{k}. One passes through P_{j}, the other through P_{i}. Note that, say P_{j}, is the center of similarity of (O_{i}) and (O_{k}). Therefore, triangles P_{j}O_{k}P_{i} and P_{j}O_{i}Q_{k} are similar. Which makes lines O_{k}P_{i} and O_{i}Q_{k} parallel. Thus, O_{k}O_{j}||O_{i}A_{k}. Similarly, O_{k}O_{i}||O_{j}A_{k}. The quadrilateral O_{i}O_{k}O_{j}A_{k} is a parallelogram. Which implies

A_{k}O_{i} = O_{j}O_{k} = r_{j} + r_{k}.

But the radius of circle (A_{k}) equals

A_{k}Q_{k} = A_{k}O_{i} + O_{i}Q_{k} = r_{j} + r_{k} + r_{i}.

Which means that the radius of circle (A_{k}) is the sum of the radii of the given three circles. Since there is nothing special about the index k, all three circles A have the same radius and are, therefore, equal.

### References

- J. Hadamard,
*Leçons de géométrie élémentaire*, tome I, 13e édition, reprint 1988, #322

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Copyright © 1996-2018 Alexander Bogomolny