What is the center and radius of the circle (x+4)^2+(y-2)^2=16

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What is the center and radius of the circle (x+4)^2+(y-2)^2=16

Mathematics
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Hint: Rewrite \[\Large (x+4)^2+(y-2)^2=16\] as \[\Large (x-(-4))^2+(y-2)^2=4^2\] and compare that to the general form\[\Large (x-h)^2+(y-k)^2=r^2\]
So the center is (4,-2) and the radius is 16
for the r what is the square root of 16?

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4
yes... so our radius is actually 4. our center is in the form of (h,k). I think opposites signs are needed. I have to recheck this one fast one.
So the center is (-4,2)
ah it's like solving for x and solving for y yeah(-4,2) is the center... radius is 4
\[\Large (x-({\color{red}{-4}}))^2+(y-{\color{blue}{2}})^2={\color{green}{4}}^2\] \[\Large (x-{\color{red}{h}})^2+(y-{\color{blue}{k}})^2={\color{green}{r}}^2\]
ok... it's (-4,2) for center radius is 4 ... whew just double checking. I"m 75% sleepy

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