5
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I want to draw a picture of $(x-t)^2 + (y-t^2)^2 - t^2$ in $xy$ plane, with different values of $t$. I did it in a very silly way , namely,

ContourPlot[{
  (x - 1)^2 + (y - 1)^2 == 1, (x - 2)^2 + (y - 4)^2 == 4, 
  (x - 3)^2 + (y - 9)^2 == 9, (x - 4)^2 + (y - 16)^2 == 16, 
  (x - 5)^2 + (y - 25)^2 == 25, (x - 6)^2 + (y - 36)^2 == 36, 
  (x - 7)^2 + (y - 49)^2 == 49, (x - 8)^2 + (y - 64)^2 == 64, 
  (x - 9)^2 + (y - 81)^2 == 81, (x - 10)^2 + (y - 100)^2 == 100, 
  (x + 1)^2 + (y - 1)^2 == 1, (x + 2)^2 + (y - 4)^2 == 4, 
  (x + 3)^2 + (y - 9)^2 == 9, (x + 4)^2 + (y - 16)^2 == 16, 
  (x + 5)^2 + (y - 25)^2 == 25, (x + 6)^2 + (y - 36)^2 == 36, 
  (x + 7)^2 + (y - 49)^2 == 49, (x + 8)^2 + (y - 64)^2 == 64, 
  (x + 9)^2 + (y - 81)^2 == 81, (x + 10)^2 + (y - 100)^2 == 100
 }, 
 {x, -110, 110}, {y, -110, 110}
]

I am sure there must be a simpler way. Could you give me a hand?

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9
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Use Table to generate the values for different t:

ContourPlot[Evaluate@Table[(x - t)^2 + (y - t^2)^2 == t^2, 
    {t, -20, 20, 1}], {x, -110, 110}, {y, -110, 110}]

enter image description here

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1
  • $\begingroup$ I missed the fact that it was $(x - t)^2 + (y - t)^2$, so the Contours option doesn't work directly, +1. $\endgroup$
    – rcollyer
    Apr 3 '12 at 15:10
6
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You can generate all formulae:

f = Function[t, (x - t)^2 + (y - t^2)^2 == t^2];
ContourPlot[Evaluate[f /@ Range[10]], {x, -110, 110}, {y, -110, 110}]

or interactively stroll through them:

Manipulate[
 ContourPlot[(x - t)^2 + (y - t^2)^2 == t^2, {x, -110, 110}, {y, -110,
    110}], {t, 1, 10, 1}]
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