Consider two intersecting circles {a(A, ra), b(B, rb)}. Draw two parallel radii AE, BF or BG from their centers. The intersection points C of lines {EF, AB} and D of {EG, AB} determine the two [similarity centers of the two circles]. D, C are [harmonic conjugate] with resepect to AB. The ratios being AD/DB = AC/BC = ra/rb.
-Extend AE to AJ. Then triangles AJC, BGC are also similar and J, G, C are on a line.
-Draw radii AI, BI of the two circles a, b respectively. From the equality of ratios AI/BI = AD/DB = AC/BC follows that DI, CI are the bisectors of the angle AIB, hence they intersect orthogonally at I, and each of the circles a, b makes equal angles with circles c and d.
-Hence the circles c(C) and d(D) are orthogonal and they [invert] both {a,b} to each other. These are called the [Mid-circles] of the two intersecting circles a and b.
-There is no other circle inverting {a,b} to each other.
EucliDraw has a tool [Circles-menu\Inversion interchanging 2 cir/lin] constructing the two circles {c,d} directly from {a,b}. One of the circles {a,b} can be a line. See the relevant variation of the figure in MidCircles2.html .
See the file MidCircle.html for the case of two non intersecting circles external to each other.
See the file Autopolar.html for an interesting illustration of the discussion to the case of obtuse triangles.
![[alogo]](alogo.png){kind=small}
Mid circles of two intersecting circles ![[0_0]](MidCircles_files/MidCircles_0_0_0.png)
![[0_1]](MidCircles_files/MidCircles_0_0_1.png)