Question

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!

Answer 1

@ganeshie8 @zepdrix @mathmate

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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.

Answer 6

Okay how do we do that ?

Answer 7

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?

Answer 8

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

Answer 9

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.

Answer 10

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

Answer 11

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.

Answer 12

(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 ?

Answer 13

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?

Answer 14

Would that be 1 also?

Answer 15

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.

Answer 16

so then that would be 2 ?

Answer 17

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

Answer 18

oh 4 ?

Answer 19

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

Answer 20

(5x)=(25x)

Answer 21

no wait typo

Answer 22

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.

Answer 23

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

Answer 24

okay so then -4

Answer 25

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

Answer 26

(3) = yk
3yk

Im sorry if thats a dumb answer i just dont know

Answer 27

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

Answer 28

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?

Answer 29

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

Answer 30

Its okay, thank you for all your help!

Answer 31

no problem! :)