Eliminating Variables in Circles Equations

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Homework Statement



I have the equations x = a * sinh(u) / (cos(v) + cosh(u)) and y = a * sin(v) / (cos(v) + cosh(u)). I believe that if a value of u is fixed, then the locus of (x,y) satisfying these equations is a circle depending on v. Similarly, I believe that if a value of v is fixed, then the locus of (x,y) satisfying these equations is a circle depending on u.

I would like to verify this.

Homework Equations





The Attempt at a Solution



It seems the easiest way is to find the equations of the circles by somehow eliminating either u or v, but I can't figure out how. I've employed various identities, such as cosh^2 - sinh^2 = 1, but I can never fully eliminate one variable.
 
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The coordinate system I'm working with is slightly different, but my conjecture about the level curves for u = const. and v = const. is based off this Wikipedia article: http://en.wikipedia.org/wiki/Bipolar_coordinates

I cannot figure out, however, how Wikipedia managed to derive the equations in the section "Curves of constant sigma and tau." I suspect a similar method will work in my case.
 
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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