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anonymous

  • 5 years ago

Its distance from (0,0) is three times its distance from (4,0)

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  1. King
    • 5 years ago
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    3,1

  2. anonymous
    • 5 years ago
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    How?

  3. King
    • 5 years ago
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    see acc. to this x=3y so if x=3 then y =1

  4. anonymous
    • 5 years ago
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    How did you get that answer? Distance formula etc

  5. amistre64
    • 5 years ago
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    id also suggest 3,0

  6. amistre64
    • 5 years ago
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    |dw:1326979014858:dw|

  7. nikvist
    • 5 years ago
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    \[x^2+y^2=(3d)^2\quad,\quad (x-4)^2+y^2=d^2\]\[x^2+y^2=9[(x-4)^2+y^2]\]\[x^2+y^2=9x^2-72x+144+9y^2\]\[0=8x^2-72x+144+8y^2\]\[0=x^2-9x+18+y^2\]\[0=x^2-9x+20.25-2.25+y^2\]\[1.5^2=(x-4.5)^2+y^2\]\[\mbox{all points on circle: }\quad 1.5^2=(x-4.5)^2+y^2\]

  8. mathmate
    • 5 years ago
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    or, for the points (0,0) and (x0,y0), all points on the circle \[(x-\frac{9x_0}{8})^2+(y-\frac{9y_0}{8})^2 = \frac{9}{64}(x_0^2+y_0^2)\] and for x0=4, y0=0, it reduces to the equation given by nikvist above. Note that both (3,0) and (6,0) satisfy this question's requirements.

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