How do you solve (x-7)^2+(y-(-11))^2=6^2

Question

anonymous · Answer

$${({x-7})^2}+{({y-(-11)}^2} = {6^2}$$

Remember your order of operations: PEMDAS.
1) Simplify the {y-(-11)}  first - remember two consecutive negatives is the same as a plus sign.
2)  You can multiply out 6^2 on the right at the same time.
3)  Lastly Use your rules to multiply a binomial by a binomial (itself).  You'll have two of these to multiply out.
4)  Finally, combine like terms.
Post your progress.

anonymous · Answer

Oops:
$$...(-11))^2$$

anonymous · Answer

Hi Denise, 
solve for what?  you have 2 variables (x and y) in the equation 
Ivan

denise · Answer

I am honestly lost!

denise · Answer

Ivan,
I am  suppose to write the center-radius form equation of the circle.

anonymous · Answer

Oops, I was telling you how to expand this equation - not very useful.
Remember the basic unit  circle equation:
$$x^2 + y^2 = (radius)^2  $$  -- radius equals 1 for the unit circle.
I imagine that your assignment is to find the new center and radius of the circle in the given equation. Correct?

anonymous · Answer

I just responded, but my answer didnt stick for some reason. I agree with Moss' interpretation. http://www.regentsprep.org/Regents/math/algtrig/ATC1/circlelesson.htm the equation you have, to me, is in the centre- radius form, whereas the multiplied out form is the "general form"