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- Denise

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

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- Denise

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

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- anonymous

\[{({x-7})^2}+{({y-(-11)}^2} = {6^2}\]
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

Oops:
\[...(-11))^2\]

- anonymous

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

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- Denise

I am honestly lost!

- Denise

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

- anonymous

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

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"

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