Solve Circle Equation: Complete the Square for Center & Radius

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To solve the equation of the circle given by y^2 + 12y = 17 - x^2 + 2x, the first step is to rearrange the equation by moving all terms except 17 to one side. This involves grouping the x and y terms separately. Completing the square for both variables is necessary to express the equation in the standard form (x-h)^2 + (y-k)^2 = r^2. The discussion highlights a misunderstanding of the completion process, emphasizing the need to correctly apply the method rather than simplifying incorrectly. Properly completing the square will reveal the center (h, k) and the radius r of the circle.
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Homework Statement



I have no idea how to figure this out. Please help!:)
Complete the square to find the center and radius of the circle:
(equation of a circle:(x-h)^2 + (y-k)^2 = r^2, where (h,k) is the center and r is the radius)

y^2+12y=17-x^2+2x

(^2 = squared)
 
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srv96 said:

Homework Statement



I have no idea how to figure this out. Please help!:)
Complete the square to find the center and radius of the circle:
(equation of a circle:(x-h)^2 + (y-k)^2 = r^2, where (h,k) is the center and r is the radius)

y^2+12y=17-x^2+2x
Move all of the terms except 17 to the left side, grouping the terms in x together and the terms in y together.

Do you know how to complete the square?
 


After completeing the square, I ended up with
y(y+12)+x(x-2)=17
Would the extra y's and x's cancel out and would the answer just become
(x-2)^2 + y(y+12)^2=square root of 17
and thanks for the previous help:)
 


srv96 said:
After completeing the square, I ended up with
y(y+12)+x(x-2)=17
Would the extra y's and x's cancel out and would the answer just become
(x-2)^2 + y(y+12)^2=square root of 17
and thanks for the previous help:)
No, that's not completing the square. Check your textbook again (if you have one). Or, here's an online lesson:
http://www.purplemath.com/modules/sqrcircle.htm"
 
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I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks

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