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!

- anonymous

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- anonymous

@ganeshie8 @zepdrix @mathmate

- mathmate

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

- mathmate

@cheetah21 Can you do that or do you need further help?

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## More answers

- anonymous

I do need further help..sorry im just really bad with math :/

- mathmate

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.

- anonymous

Okay how do we do that ?

- mathmate

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?

- anonymous

x2 + 2x + y2 + 4y = 20
(x2 + 2x) + (y2 + 4y) = 20

- mathmate

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.

- anonymous

Oh okay im sorry , i was confused i guess :/ so now do we need to solve in the parentheses or what ?

- mathmate

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.

- anonymous

(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 ?

- mathmate

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?

- anonymous

Would that be 1 also?

- mathmate

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.

- anonymous

so then that would be 2 ?

- mathmate

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

- anonymous

oh 4 ?

- mathmate

(x+1)^2 = (x-4)^2 (are they equal?)

- anonymous

(5x)=(25x)

- anonymous

no wait typo

- mathmate

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.

- mathmate

Can you try finding k from:
(y+2)^2 = (y-k)^2 ?

- anonymous

okay so then -4

- mathmate

You would take the square root on both sides, so
(y+2) = y-k,
can you solve for k?

- anonymous

(3) = yk
3yk
Im sorry if thats a dumb answer i just dont know

- mathmate

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

- anonymous

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?

- mathmate

put it up, and perhaps someone else will look at it for you. I have to go now, sorry! :(

- anonymous

Its okay, thank you for all your help!

- mathmate

no problem! :)

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