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Math Help - Find the center and radius of a circle?

  1. #1
    Member M670's Avatar
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    Find the center and radius of a circle?

    Find the coordinates of the center and the radius of the circle given by the equation  y(y-4)+x(x+3)=0
    I know  (h,k) is the center which if I am right its (4,-3) but how do I find the radius ?
    if I take  y(y-4)+x(x+3)=0 and expand it out to y^2-4y+x^2+3x=0 how do I enter that into (x-h)^2+(y-k)^2=r^2 Because I think I need to find atleast one coordinate to find my radius?
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  2. #2
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    Re: Find the center and radius of a circle?

    Quote Originally Posted by M670 View Post
    Find the coordinates of the center and the radius of the circle given by the equation  y(y-4)+x(x+3)=0
    I know  (h,k) is the center which if I am right its (4,-3) but how do I find the radius ?
    if I take  y(y-4)+x(x+3)=0 and expand it out to y^2-4y+x^2+3x=0 how do I enter that into (x-h)^2+(y-k)^2=r^2 Because I think I need to find atleast one coordinate to find my radius?
    1. You've done the first step:

     y(y-4)+x(x+3)=0 expands to y^2-4y+x^2+3x=0

    2. Now complete the square(s):

    y^2-4y+4+x^2+3x+\left(\frac32 \right)^2=4+\left(\frac32 \right)^2

    \left(x+\frac32 \right)^2 + \left(y-2 \right)^2=\frac{25}4 = \left(\frac52 \right)^2

    3. Thus the midpoint of the circle is M\left(-\frac32 , 2 \right) and the radius is r = \frac52
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  3. #3
    Member M670's Avatar
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    Re: Find the center and radius of a circle?

    Quote Originally Posted by earboth View Post
    1. You've done the first step:

     y(y-4)+x(x+3)=0 expands to y^2-4y+x^2+3x=0

    2. Now complete the square(s):

    y^2-4y+4+x^2+3x+\left(\frac32 \right)^2=4+\left(\frac32 \right)^2

    \left(x+\frac32 \right)^2 + \left(y-2 \right)^2=\frac{25}4 = \left(\frac52 \right)^2

    3. Thus the midpoint of the circle is M\left(-\frac32 , 2 \right) and the radius is r = \frac52

    Thank You, I was missing my step of completing the squares
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