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anonymous

  • 5 years ago

Help!!! Find the differential equation of all the circles with centers x + 2y = 0 and passing through (0,0).

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  1. anonymous
    • 5 years ago
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    How about this... A circle with center (-2u, u), radius r is written as \[(x + 2u)^{2} + (y - u)^{2} = r^{2}\] Differentiating both side with respect to x, we get \[ x + 2u + (y - u)y' = 0\]

  2. anonymous
    • 5 years ago
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    how do you get the center?

  3. anonymous
    • 5 years ago
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    This circle \[ (x+2u)^2+(y−u^)2=r^2\] has its centre at (-2u, u), right?

  4. anonymous
    • 5 years ago
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    the centers of the all circles are lying at a line x + 2y = 0. how do you get the center (-2u,u) and the circle must passing through (0,0).

  5. anonymous
    • 5 years ago
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    Let x = -2u and y = u. For any u, (-2u) + 2u = 0. So (-2u, u) is on the line x + 2y = 0. I got this pair (-2u, u) just by modifying x + 2y = 0 -> x = -2y.

  6. anonymous
    • 5 years ago
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    ok, thanks

  7. anonymous
    • 5 years ago
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    I'm glad if it helped. Thanks.

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