## Mid circle of two non intersecting circles

Here handled the case of two disjoint circles a and b, b being internal to a. For the external case see MidCircle.html .

Consider two non intersecting circles {a(A, ra), b(B, rb)}. Draw two parallel radii AE, BF from their centers. The intersection point C of lines {EF, AB} determine the internal [similarity center of the two circles]. The ratio being AC/BC = ra/rb.
-Extend BF defining H. Then triangle HGF is isosceles and circle h(H) is tangent to both a and b.
-Define circle c(C) by the property to be centered at C and orthogonal to h(H). The inversion w.r. to c interchanges {a,b}.
-The radius rc of c(C) is independent of the particular h(H). c(C) is orthogonal to every circle tangent to both {a,b}, the tangency being external for b and internal for a.
-c(C) is also orthogonal to every circle d(D) orthogonal to both {a,b}. Hence it belongs to the coaxal system of {a,b}.
-c(C) is called the [Mid-circle] of a and b. See the file MidCircles.html for other configurations of the two circles.

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