I have a general question about drawing the reciprocals of circles through the circle of reciprocation.
I understand inversion and reciprocation are two entirely different things yet somehow connected.
When reciprocating a circle(alpha) outside of the circle of reciprocation(O) , do we draw tangent lines on alpha, then draw respective pependiculars through O and call that point of intersection it's reciprocal?
What difference did you understand between inversion and reciprocation? Perhaps the calculation of polar coordinate radius is reciprocation?
When a circle of radius $r_1$ is reflected / mirrored/ inverted about a circle of inversion of radius $a$ with center at origin we have in polar coordinates
$$ r_2 \rightarrow a^2/r_1 ;\ \rho_2 \rightarrow a^2/\rho_1$$
for polar coordinate and circle radius, and,
$$ x_2 = \frac{x_1 a^2 }{(x_1^2+y_1^2)} ; \, y_2 = \frac{y_1 a^2}{(x_1^2+y_1^2)} ; $$
in rectangular coordinates.
Perpendiculars in construction are tangency point as shown. At corresponding points of maximum, points of tangency and minimum radius note that
$$ r_{1min} \cdot r_{2 max} = r_{1max} \cdot r_{2 min }= r_{\alpha @ tangent }\cdot r_{\beta @ tangent } = a^2 $$
Please see the similarity of triangles in the image, corresponding sides are $ PA, P^{\prime}B^{\prime}$ etc. in Wu's site below.
May be better to avoid the term reciprocation in this context, as inversion serves the same purpose.
Wu Circle inversion