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anonymous

  • one year ago

Graph the equation with a diameter that has endpoints at (-2, -2) and (4, 2). Label the center and at least four points on the circle. Write the equation of the circle.

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  1. anonymous
    • one year ago
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    @jdoe0001

  2. anonymous
    • one year ago
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    Heey, this is kinda the same question as befre, but I don't think my answer for distance is correct

  3. anonymous
    • one year ago
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    @Directrix

  4. jdoe0001
    • one year ago
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    hmm what did you get for the distance?

  5. anonymous
    • one year ago
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    About 7.21

  6. jdoe0001
    • one year ago
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    hmm is there a radical form for it?

  7. anonymous
    • one year ago
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    Sorry...?

  8. jdoe0001
    • one year ago
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    well. you used the distance formula to get it.... so... that means you'd have a root of it something like \(\bf distance = \sqrt{\qquad}\)

  9. anonymous
    • one year ago
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    \[Distance= \sqrt{(-2-4)^2+(-2-2)^2}\]

  10. jdoe0001
    • one year ago
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    yeap... I got the same, which in decimal form is about 7.21 so, that's correct \(\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({\color{red}{ -2}}\quad ,&{\color{blue}{ -2}})\quad % (c,d) &({\color{red}{ 4}}\quad ,&{\color{blue}{ 2}}) \end{array}\qquad % distance value d = \sqrt{6^2+4^2}\implies d=\sqrt{52}\impliedby diameter \\ \quad \\ radius=\cfrac{\sqrt{52}}{2}\)

  11. anonymous
    • one year ago
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    So the radius would be 3.60

  12. jdoe0001
    • one year ago
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    using the midpoint formula, the center is at 1,0 thus \(\large (x-{\color{brown}{ h}})^2+(y-{\color{blue}{ k}})^2={\color{purple}{ r}}^2 \qquad center\ ({\color{brown}{ 1}},{\color{blue}{ 0}})\qquad radius={\color{purple}{ \frac{\sqrt{52}}{2}}}\)

  13. anonymous
    • one year ago
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    Wouldn't the center be (1,1)

  14. anonymous
    • one year ago
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    ?

  15. jdoe0001
    • one year ago
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    lemme recheck that

  16. anonymous
    • one year ago
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    Oops, nevermind, my bad, it is 0

  17. jdoe0001
    • one year ago
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    yeah.. is 1,0

  18. jdoe0001
    • one year ago
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    shoot.. a bit truncated...lemme redo that

  19. jdoe0001
    • one year ago
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    \(\bf (x-{\color{brown}{ h}})^2+(y-{\color{blue}{ k}})^2={\color{purple}{ r}}^2 \qquad center\ ({\color{brown}{ 1}},{\color{blue}{ 0}})\qquad radius={\color{purple}{ \frac{\sqrt{52}}{2}}} \\ \quad \\ \quad \\ (x-{\color{brown}{ 1}})^2+(y-{\color{blue}{ 0}})^2=\left( {\color{purple}{ \frac{\sqrt{52}}{2}}}\right)^2 \\ \quad \\ (x-1)^2+y=\cfrac{(\sqrt{52})^2}{2^2}\implies (x-1)^2+y=\cfrac{\cancel{52}}{\cancel{4}} \\ \quad \\ \implies (x-1)^2+y=13\)

  20. anonymous
    • one year ago
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    So the square root is cancelled off in the equation but I still have a radius of 3.60 right?

  21. jdoe0001
    • one year ago
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    if you want the decimal form of it, yes the advantage of the radical form is that, it doesn't have any missing floating values

  22. anonymous
    • one year ago
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    Yeah, but I need the decimal so that I can graph it

  23. jdoe0001
    • one year ago
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    but yes, 7.21 is a rounded up version, you lose a few decimals there half that is 3.6, also rounded up, again losing a few decimals

  24. jdoe0001
    • one year ago
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    welll. yes sure.. but yes, is 3.6 :)

  25. anonymous
    • one year ago
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    Thank u!!

  26. triciaal
    • one year ago
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    |dw:1437188896778:dw|

  27. triciaal
    • one year ago
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    my center is (1, -1)

  28. triciaal
    • one year ago
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    radius is 1/2 the diameter endpoints (-2,2) and ( 4,2) diameter = 6 units so radius = 3 units for the x value -2 +3 or 4 -3 = 1 from end to the center = 3 units 2 -3 = -1 so the y value for the center = -1

  29. triciaal
    • one year ago
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    oops I did -2, 2 should be -2, -2 sorry

  30. triciaal
    • one year ago
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    |dw:1437190476623:dw|

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