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Math Help - Find the center and radius of circle

  1. #1
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    Find the center and radius of circle

    Question : Find the center and radius of the circle

    2x^2 + 2y^2 + 5x + 7y -3 = 0

    My Work:::::::::::

    center C(- \frac{5}{4} , - \frac{7}{2})..............Is this correct?

    Raduis = \frac{7\sqrt{5}}{4}......................Is this correct?
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  2. #2
    Super Member bigwave's Avatar
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    did you change this expression to to

     <br />
(x-h)^2 + (y-k)^2 = r^2<br />

    center is (h,k) and radius r>0
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  3. #3
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    No !!!!!

    just used the general equation and found the center and raduis through it?????
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  4. #4
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    Quote Originally Posted by zorro View Post
    Question : Find the center and radius of the circle

    2x^2 + 2y^2 + 5x + 7y -3 = 0

    My Work:::::::::::

    center C(- \frac{5}{4} , - \frac{7}{2})..............Is this correct?

    Raduis = \frac{7\sqrt{5}}{4}......................Is this correct?
    You need to complete the square for x and y.

    2x^2 + 2y^2 + 5x + 7y - 3 = 0

    2x^2 + 5x + 2y^2 + 7y = 3

    x^2 + \frac{5}{2}x + y^2 + \frac{7}{2}y = \frac{3}{2}

    x^2 + \frac{5}{2}x + \left(\frac{5}{4}\right)^2 + y^2 + \frac{7}{2}y + \left(\frac{7}{4}\right)^2 = \frac{3}{2} + \left(\frac{5}{4}\right)^2 + \left(\frac{7}{4}\right)^2

    \left(x + \frac{5}{4}\right)^2 + \left(y + \frac{7}{4}\right)^2 = \frac{49}{8}

    \left[x - \left(-\frac{5}{4}\right)\right]^2 + \left[y - \left(-\frac{7}{4}\right)\right]^2 = \left(\sqrt{\frac{49}{8}}\right)^2

    \left[x - \left(-\frac{5}{4}\right)\right]^2 + \left[y - \left(-\frac{7}{4}\right)\right]^2 = \left(\frac{7}{2\sqrt{2}}\right)^2

    \left[x - \left(-\frac{5}{4}\right)\right]^2 + \left[y - \left(-\frac{7}{4}\right)\right]^2 = \left(\frac{7\sqrt{2}}{4}\right)^2.


    Can you read off the centre and the radius now?
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