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 one [similarity center of the two circles]. The ratio being AC/BC = ra/rb.

-Extend AG and BF defining H. Then triangle HGF is isosceles and circle h(H) is tangent externally 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} externally or internally (both).

-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.

See the file MidCircle2.html for the case of two circles tangent at a point.

Notice that C is the point dividing AB in ratio ra/rb externally, whereas if b is non intersecting a but internal to it then C must be taken internal to AB. See the relevant figure in MidCircle3.html .

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