Given the following:
x^2+y^2-6y+4x-16=0
how do I find the measure of the radius?
$\displaystyle
x^2+y^2-6y+4x-16=0
$
$\displaystyle
(x+2)^2+(y-3)^2=29
$
Obviously, the radiius is $\displaystyle \sqrt{29}$
ALITER:
The general equation of a circle is
$\displaystyle
x^2+y^2+2gx+2fy+c=0
$
the centre being $\displaystyle (-g,-f)$ and the $\displaystyle radius=\sqrt{g^2+f^2-c}$
As per your question $\displaystyle g=2,f=-3,c=-16$
Therefore
$\displaystyle
radius=\sqrt{2^2+(-3)^2-(-16)}=\sqrt{29}
$
Are you familiar with completing the square?
$\displaystyle x^{2} + y^{2} - 6y + 4x - 16 = 0$
Rearrange terms:
$\displaystyle x^{2} + 4x + y^{2} - 6y = 16$
Complete the square:
$\displaystyle x^{2} + 4x + 4 + y^{2} - 6y + 9 = 16 + 4 + 9$
Factor and simplify:
$\displaystyle (x + 2)^{2} + (y - 3)^{2} = 29$
01
You can "create" anything in one side of the equation as long as your create the same thing on the other side. So, 4 and 9 were added to the left side of the equation so that the square could be completed, but they had to also be added to the right side to balance things out.