Using the following equation, find the center and radius of the circle. You must show all work and calculations to receive credit. x2 + 2x + y2 + 4y = 20 I am completely lost on how to do this!!! I will fan and medal please someone help!

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Using the following equation, find the center and radius of the circle. You must show all work and calculations to receive credit. x2 + 2x + y2 + 4y = 20 I am completely lost on how to do this!!! I will fan and medal please someone help!

Mathematics
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Complete the square for x and y, then you'll get the equation of the circle with centre (h,k) and radius r as in (x-h)^2+(y-k)^2 = r^2
@cheetah21 Can you do that or do you need further help?

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I do need further help..sorry im just really bad with math :/
Start with the given equation, we need to transform it to the standard form of the circle, then we can identify (by comparison) h,k and r.
Okay how do we do that ?
The given equation is: x2 + 2x + y2 + 4y = 20 After adding a 1 and a 4 on both sides, we put the equation as (x^2+2x+1) + (y^2+2(2y)+4) = 20 +1 +4 It's still the same equation, but what are in the parentheses are perfect squares. Can you factor the two expressions in parentheses?
x2 + 2x + y2 + 4y = 20 (x2 + 2x) + (y2 + 4y) = 20
I already put them in perfect squares, you don't need to simplify them. In fact: (x^2+2x+1) + (y^2+2(2y)+4) = 20 +1 +4 is the same as: (x+1)^2 + (y+2)^2 = 25 You can check to make sure by expanding the squares.
Oh okay im sorry , i was confused i guess :/ so now do we need to solve in the parentheses or what ?
Next step is easy: you compare (x+1)^2 + (y+2)^2 = 5^2 with the standard form of the circle with centre (h,k) and radius r (x-h)^2+(y-k)^2 = r^2 and decide what the values of h, k and r are.
(X-h)^2 + (y - K)^2 = r^2 (x-h)^2+(y-k)^2 = r^2 is that what it should look like right now ?? and how do we figure out values ?
Compare term by term, for example, (x+1)^2 = (x-h)^2, what would be the value of h to make the right hand side (x+1)^2?
Would that be 1 also?
If h=1, then we have (x+1)^2 on the left hand side, and (x-1)^2 , so the two sides don't equal. You need to find another value of h.
so then that would be 2 ?
If h=2, then we have (x+1)^2 on the left hand side, and (x-2)^2 , so the two sides don't equal. Remember we are looking for h in: (x+1)^2 = (x-h)^2
oh 4 ?
(x+1)^2 = (x-4)^2 (are they equal?)
(5x)=(25x)
no wait typo
There is a positive sign on the left, and a negative sign on the right, so h must be negative to make the two sides match, remembering -(-1)=+1. so putting h=-1 will make (x+1)^2 = (x-(-1))^2=(x+1)^2 or both sides equal.
Can you try finding k from: (y+2)^2 = (y-k)^2 ?
okay so then -4
You would take the square root on both sides, so (y+2) = y-k, can you solve for k?
(3) = yk 3yk Im sorry if thats a dumb answer i just dont know
From (y+2) = y-k, subtract y from both sides, so y-y+2 = y-y-k or 2=-k, multiply by -1 to get -2=k, so we have h=-1, k=-2, and r=5 From the standard equation of a circle, (x-h)^2+(y-k)^2=r^2 we therefore found centre = (h,k)=-1,-2, and radius = r =5
wait so thats the answer to the eqaution ? oh okay , well if i put all the stuff we have come up with together can you tell me if its in the correct order?
put it up, and perhaps someone else will look at it for you. I have to go now, sorry! :(
Its okay, thank you for all your help!
no problem! :)

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