Find the center and radius of a circle?

**M670** · December 1st 2012, 10:21 AM

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?

**earboth** · December 1st 2012, 11:01 AM

Originally Posted by M670

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 $r = \frac52$

**M670** · December 1st 2012, 11:10 AM

Originally Posted by earboth

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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