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moongazer

  • 4 years ago

Find the radius and the center of the circle. Just check my answer.

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  1. amistre64
    • 4 years ago
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    (x-Cx)^2+(y-Cy)^2=r^2

  2. moongazer
    • 4 years ago
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    2x^2 + 2y^2 - 2x + 2y + 7 = 0 My answer is : c(1/2,-1/2) r= \[\sqrt{-7/2}\]

  3. moongazer
    • 4 years ago
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    Am I right? Cause I am having doubt that my answer is wrong because I have a negative radius

  4. amistre64
    • 4 years ago
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    2x^2 + 2y^2 - 2x + 2y + 7 = 0 2x^2 - 2x + 2y^2 + 2y + 7 = 0 2(x^2 - x) + 2(y^2 +y) = -7 2(x^2 - x +(1/2)^2) + 2(y^2 +y+(1/2)^2) = -7 + 1 2(x-1/2)^2 + 2(y+1/2) = -6 its quite possible that this cannot produce a real circle; just becasue something has some numbers thrown about doesnt make it sensible :) lets chk the wolf

  5. amistre64
    • 4 years ago
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    and make sure that aint spose to be a -7 instead

  6. amistre64
    • 4 years ago
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    http://www.wolframalpha.com/input/?i=2x%5E2+%2B+2y%5E2+-+2x+%2B+2y+%2B+7+%3D+0 as is its right, but no circle

  7. mathmate
    • 4 years ago
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    Check for typos: Could the question be 2x^2 + 2y^2 - 2x + 2y = 7 or 2x^2 + 2y^2 - 2x + 2y - 7 = 0

  8. moongazer
    • 4 years ago
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    That's why I am having doubts with my answer. :) Maybe I just copied the text wrong.

  9. moongazer
    • 4 years ago
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    If it is -7 will the answers be c(1/2,-1/2) r=\[\sqrt{7/2}\]

  10. moongazer
    • 4 years ago
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    ??

  11. amistre64
    • 4 years ago
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    if -7 then we get r^2 = 8 center: (1/2, 1/2)

  12. amistre64
    • 4 years ago
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    -1/2 yeah, but r=sqrt(8) = 2sqrt(2)

  13. amistre64
    • 4 years ago
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    did that wrong somehow, radius is 2 according to the wolf

  14. amistre64
    • 4 years ago
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    http://www.wolframalpha.com/input/?i=2x%5E2+%2B+2y%5E2+-+2x+%2B+2y+-+7+%3D+0

  15. moongazer
    • 4 years ago
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    r= (7sqrt(2))/2 ??

  16. amistre64
    • 4 years ago
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    \[r^2=7+2(1/2)^2+2(1/2)^2\]

  17. moongazer
    • 4 years ago
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    how did it became x = -3/2, x = 5/2 ???

  18. amistre64
    • 4 years ago
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    i see what i did; we divide off the 2 and get 8/2 = 4 sqrt(4) = 2

  19. amistre64
    • 4 years ago
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    your responses arent making sense to me

  20. amistre64
    • 4 years ago
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    \[2(x^2 - x +(1/2)^2) + 2(y^2 +y+(1/2)^2) = 7 + 1\] \[2(x-(1/2))^2 + 2(y+(1/2))^2 = 8\] \[(x-(1/2))^2 + (y+(1/2))^2 = 4\]

  21. amistre64
    • 4 years ago
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    those are points on the circle that the wolf handed out

  22. moongazer
    • 4 years ago
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    2(x^2 - x +(1/2)^2) + 2(y^2 +y+(1/2)^2) = 7 + 1 Why did you add one on the right side of the equation?

  23. amistre64
    • 4 years ago
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    becasue we add a total of 1 to the left side 2(1/2)^2 + 2(1/2)^2 = 1

  24. amistre64
    • 4 years ago
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    2x^2 - 2x +2(1/2)^2 + 2y^2 +2y +2(1/2)^2) = 7 + 1 might be easier to see that way

  25. amistre64
    • 4 years ago
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    i dont divide off the "2" right away which might be a difference in our computing techniques

  26. moongazer
    • 4 years ago
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    ohh ok, I forgot adding the numbers that I added on the left side to the right side. That's why I got confused. :) Anyway, I divided the equation directly to "2" but I still got the right answer. :)

  27. moongazer
    • 4 years ago
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    thanks!

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