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The centers of all circles which tangent to a circle $P(2R,0)$ with radius $R$ and to the line $x=-t$ is a Canonical parabola. Need to find the equation of that parabola and $t$ (by $R$).

so: \begin{align}&\sqrt{{{(x-2R)}^{2}}+{{y}^{2}}}=R+\left| x+t\right|\\ &\Leftrightarrow{{(x-2R)}^{2}}+{{y}^{2}}={{x}^{2}}+2tx+{{t}^{2}}+{{R}^{2}}+2R\left| x+t\right|\\ &\Leftrightarrow{{x}^{2}}+4{{R}^{2}}-4Rx+{{y}^{2}}={{x}^{2}}+2tx+{{t}^{2}}+{{R}^{2}}+2R\left| x+t\right|\end{align}

but how can i get to the answers:

$t=R$, $y^2=8RX$?

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  • $\begingroup$ Can you post the equation of the circle $P(2R,0)$? $\endgroup$ Jul 2, 2013 at 17:29
  • $\begingroup$ Does all other circles have radius $R$ or it can vary? $\endgroup$ Jul 2, 2013 at 17:29

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