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:<math>x \to \left [ (c-a)^{-1} - (b-a)^{-1} \right ]^{-1} x</math>
Then <math>a</math> stays at <math>\infty</math>, <math>b \to 0</math>, <math>c \to 1</math>, <math>d \to z</math> (say). The unique semicircle, with center at the origin, perpendicular to the one on <math>1z</math> must have a radius tangent to the radius of the other. The right triangle formed by the abscissa and the perpendicular radii has [[hypotenuse]] of length <math>\begin{matrix} \frac{1}{2} \end{matrix} (z+1)</math>. Since <math>\begin{matrix} \frac{1}{2} \end{matrix} (z-1)</math> is the radius of the semicircle on <math>1z</math>, the common perpendicular sought has radius-square
:<math>\frac{1}{4} \left [ (z+1)^2 - (z-1)^2 \right ] = z.</math>
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* If both chords are diameters, they intersect.(at the center of the boundary circle)
* If only one of the chords is a diameter, the other chord projects orthogonally down to a section of the first chord contained in its interior, and a line from the pole orthogonal to the diameter intersects both the diameter and the chord.
* If both lines are not diameters, then we may extend the tangents drawn from each pole to produce a [[quadrilateral]] with the [[unit circle]] inscribed within it.{{how|date=August 2015}} The poles are opposite vertices of this quadrilateral, and the chords are lines drawn between adjacent sides of the vertex, across opposite corners. Since the quadrilateral is convex,{{why|date=August 2015}} the line between the poles intersects both of the chords drawn across the corners, and the segment of the line between the chords defines the required chord perpendicular to the two other chords.
<!-- ??? "then we may extend the tangents drawn from each pole to produce a [[quadrilateral]] with the unit circle inscribed within it " this is not always the case, they not always form a quadrilateral, nor is the quadrilateral always convex see also http://math.stackexchange.com/q/1382739/88985 -->
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