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Math Help - equation of circle

  1. #1
    Junior Member R3ap3r's Avatar
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    equation of circle

    Find the equation of the circle with center at (-2, 3) and which passes through (-1, 4).

    My answer:

    (x+1)^2 + (y-3)^2=1

    Hows it look?
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  2. #2
    Bar0n janvdl's Avatar
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    Quote Originally Posted by R3ap3r View Post
    Find the equation of the circle with center at (-2, 3) and which passes through (-1, 4).

    My answer:

    (x+1)^2 + (y-3)^2=1

    Hows it look?
    Now this I will not get wrong...

    (x + 2)^2 + (y - 3)^2 = r^2

    (-1 + 2)^2 + (4-3)^2 = r^2

    r^2 = 2

    r = \sqrt{2}
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  3. #3
    Moo
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    A Cute Angle Moo's Avatar
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    Hello,

    Quote Originally Posted by R3ap3r View Post
    Find the equation of the circle with center at (-2, 3) and which passes through (-1, 4).

    My answer:

    (x+1)^2 + (y-3)^2=1

    Hows it look?
    Hmm...not so good :/

    The general equation of a circle is (x-x_O)^2+(y-y_O)^2=R^2

    Where the center is O ( x_O;y_O). Here, the center is (-2,3).

    For the radius, replace : you know that (-1,4) is on the circle. So replace x=-1 and y=4 to get Rē.
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  4. #4
    Junior Member R3ap3r's Avatar
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    lol my eye sight fails.
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  5. #5
    Junior Member R3ap3r's Avatar
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    (x+2)^2 + (y-3)^2 = 2

    You think I should write it as 2 or (\sqrt{2})^2
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  6. #6
    Bar0n janvdl's Avatar
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    Quote Originally Posted by R3ap3r View Post
    (x+2)^2 + (y-3)^2 = 2

    You think I should write it as 2 or (\sqrt{2})^2
    Keep it simple...
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  7. #7
    Moo
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    Quote Originally Posted by R3ap3r View Post
    (x+2)^2 + (y-3)^2 = 2

    You think I should write it as 2 or (\sqrt{2})^2
    It depends on how you want things to be clear.

    The equation of the circle will be the first one.

    On your rough copy, it'll be the second one, so that you know that the radius is \sqrt{2}

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  8. #8
    Junior Member R3ap3r's Avatar
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    gotcha
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