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anonymous

  • one year ago

The equation of a circle is in the form x^2 +y^2 +Cx+Dy+E=. Find an equation of the circle shown in the illustration by determining C,D, and E

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  1. Michele_Laino
    • one year ago
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    where is the illustration, please?

  2. anonymous
    • one year ago
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    |dw:1444143802955:dw|

  3. Michele_Laino
    • one year ago
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    here as we can see the center of your circumference is at point \((3,0)\)

  4. Michele_Laino
    • one year ago
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    sorry, we have to replace the coordinates of the points you provided, into your generic equation

  5. anonymous
    • one year ago
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    How? I'm so confused

  6. Michele_Laino
    • one year ago
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    for examnple, your circumference passes at point \((0,0)\) so I replace \(x=0,y=0\) into your equation, so I get this: \[{0^2} + {0^2} + C \cdot 0 + D \cdot 0 + E = 0\] what is \(E\)?

  7. anonymous
    • one year ago
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    Even plug in the points (3,3) and (6,0)?

  8. Michele_Laino
    • one year ago
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    that's right!

  9. anonymous
    • one year ago
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    okay thank you so much

  10. Michele_Laino
    • one year ago
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    :)

  11. anonymous
    • one year ago
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    Wait @Michele_Laino

  12. anonymous
    • one year ago
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    I have to write and solve a system of three equations in the three unknowns C, D, and E. Then write the equation of the circle in general form:

  13. anonymous
    • one year ago
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    wait @Michele_Laino

  14. Michele_Laino
    • one year ago
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    yes! correct! As you can see we have an algebraic system with 2 unknowns, since we have already determined E as \(E=0\)

  15. anonymous
    • one year ago
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    But don't we have to come up with 3 systems

  16. Michele_Laino
    • one year ago
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    if we replace \(x=3,y=3\) into your equation,we get: 9+9+3C+3D=0 that is the first equation of our system, the first one is E=0

  17. Michele_Laino
    • one year ago
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    if we replace \(x=6,y=0\), we get: 36+0+6C+0.D=0 that is the third equation

  18. Michele_Laino
    • one year ago
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    please solve those new equations with respect to C and D

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