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Thread: 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 $\displaystyle y(y-4)+x(x+3)=0$
    I know $\displaystyle (h,k)$ is the center which if I am right its $\displaystyle (4,-3)$ but how do I find the radius ?
    if I take $\displaystyle y(y-4)+x(x+3)=0$ and expand it out to $\displaystyle y^2-4y+x^2+3x=0$ how do I enter that into $\displaystyle (x-h)^2+(y-k)^2=r^2$ 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 $\displaystyle y(y-4)+x(x+3)=0$
    I know $\displaystyle (h,k)$ is the center which if I am right its $\displaystyle (4,-3)$ but how do I find the radius ?
    if I take $\displaystyle y(y-4)+x(x+3)=0$ and expand it out to $\displaystyle y^2-4y+x^2+3x=0$ how do I enter that into $\displaystyle (x-h)^2+(y-k)^2=r^2$ Because I think I need to find atleast one coordinate to find my radius?
    1. You've done the first step:

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

    2. Now complete the square(s):

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

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

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

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

    2. Now complete the square(s):

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

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

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

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