Prove Analytically: Inversion of a Circle is Also a Circle

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Homework Statement



Given the unit circle (in the Euclidean plane) centered at the origin x^2+y^2=1, and a general circle D with equation (x-a)^2+(y-b)^2=c^2 that does not pass through the origin (ie the center of inversion, ie a^2+b^2≠c^2, prove analytically that the inversion of D in the unit cirlce is still a circle.

Homework Equations



See the attached pdf files

The Attempt at a Solution



I can prove this synthetically. I even worked out the equation of the image circle D', but I can't derive it algebraically. I feel I must be missing something very obvious.

I uploaded two pdf files. I was going to upload a GSP file, but I guess this forum can't do that? I'll have to generate pdfs from it or something. Do most of you guys have GSP? It's mind-bogglingly useful.
 

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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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