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Suppose the following function $f(t)$

$$f(t)=x^2+y^2+2m(u-y)-u^2,$$

where $$m=\frac{(t-2)(1-t^{2}+\sqrt{1-t^{2}})}{t(1+\sqrt{1-t^{2}})(1-t-\sqrt{1-t^{2}})},$$

and

$$u=\frac{t}{1+\sqrt{1-t^{2}}}.$$

Is there any set of methods in Mathematica cancelling the denominator and rearranging the terms by powers of $t$ in the following equation

$$f(t)=0,$$

that probably leads to the quartic equation (I am not sure)?

The following code:

m = (t - 2)*(1 - t^2 + Sqrt[1 - t^2])/(t*(1 + Sqrt[1 - t^2]))*(1 - t - Sqrt[1 - t^2])
u = t/(1 + Sqrt[1 - t^2])
FullSimplify[x^2 + y^2 + 2*m*(u - y) - u^2 == 0]

leads to

$$\frac{2 (t-2) \left(\sqrt{1-t^2}+t-1\right) \left(-t^2+\sqrt{1-t^2}+1\right) \left(\sqrt{1-t^2} y-t+y\right)}{t \left(\sqrt{1-t^2}+1\right)^2}+x^2+y^2=\frac{t^2}{\left(\sqrt{1-t^2}+1\right)^2},$$

that looks much better.

Thanks for your help.

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    $\begingroup$ FullSimplify[Together[x^2 + y^2 + 2*m*(u - y) - u^2] == 0]? $\endgroup$
    – kglr
    Sep 11, 2018 at 13:02
  • $\begingroup$ @ kglr: Thanks, the denominator has been removed. $\endgroup$
    – justik
    Sep 11, 2018 at 13:34

1 Answer 1

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Amplifying on kglr's comment

m = (t - 2)*(1 - t^2 + Sqrt[1 - t^2])/(t*(1 + Sqrt[1 - t^2]))*(1 - t - 
     Sqrt[1 - t^2]);
u = t/(1 + Sqrt[1 - t^2]);

If you can assume t <= 1 and all the variables are real,

eqn1 = x^2 + y^2 + 2*m*(u - y) - u^2 == 0 // Together // FullSimplify //
   ComplexExpand[#, TargetFunctions -> {Re, Im}] & // 
  FullSimplify[#, t <= 1] &

(* 3 + x^2 + 2 t (-1 + y) + 4 t^2 (-1 + y) + y^2 + 
  Sqrt[1 - t^2] (-3 + x^2 + (-4 + y) y + 6 t (1 + y) - 2 t^2 (1 + y)) == 
 2 t^3 (-1 + y) + 4 y *)

Or if you prefer the form f[t] == 0

eqn2 = SubtractSides[eqn1, eqn1[[-1]]]

(* 3 + x^2 + 2 t (-1 + y) + 4 t^2 (-1 + y) - 2 t^3 (-1 + y) - 4 y + y^2 + 
  Sqrt[1 - t^2] (-3 + x^2 + (-4 + y) y + 6 t (1 + y) - 2 t^2 (1 + y)) == 0 *)
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