Question

Using the following equation, find the center and radius of the circle you must show all work and calculation to recieve credit

x^2 -4x+y^2+8y=-4

Answer 1

I need help urgent

Answer 2

Are you familiar with the concept of completing the square?

Answer 3

Kind of

Answer 4

Do you know the general form of the equation of a circle with radius r and center (a, b)?

Answer 5

Yes

Answer 6

So you know you have to rewrite the given equation to the form:

(x-a)²+(y-b)²=r².

First, try to complete the square in x² - 4x.

Answer 7

Sorry, that looks like incomprehensible stuff!

Answer 8

Sorry i am on a ipad and cannot see that its just a bunch of squared

Answer 9

So you know you have to rewrite the given equation to the form:

(x-a)^2+(y-b)^2=r².

First, try to complete the square in x^2 - 4x.

Answer 10

It should read r^2 after the =

Answer 11

I do not understand how would i complete theat part of the equation

Answer 12

you could write x^2-4x = (x-2)^2 - 4, because, if you expand this, you''l get:

x^2-4x+4-4, which is what you had at the beginning. This is called "completing the square".

So, in your equation, you can replace the x^2-4x with : (x-2)^2 - 4.

Answer 13

It has to do with this well-known rule: (a - b)^2 = a^2 - 2ab + b^2

Answer 14

Ok how do we continue

Answer 15

Do the same with y^2 + 8y

Answer 16

(Y-2)^4???

Answer 17

No, just make a complete square!

If you take (y +4)^2 and expand it, you have: y^2 + 8y + 16.
This is almost what is in your equation: there it reads: y^2 + 8y.

So if you would replace y^2 + 8y with (y + 4)^2, it would be 16 to big.
Well, then replace y^2 +8y with (y + 4)^2 - 16, then everybody is happy.
We have completed the square.

Answer 18

Do you understand what I did there?

Answer 19

Yes

Answer 20

Thank you

Answer 21

OK, fine!
Now let us see what we have:

x^2 -4x+y^2+8y=-4 was the original equation.

we replaced x^2 - 4x with (x - 2)^2 -4.
we replaced y^2 + 8y with (y +4)^2 - 16.

This leaves us with:

(x-2)^2 - 4 + (y + 4)^2 - 16 = -4.

Now for the last step: convert to the form:

(x - a)^2 + (y - b)^2 = r^2.

Can you do this?

Answer 22

Not sure but (x+6)^2+(y-12)^2=r^2

Answer 23

You are confusing the squares and the constants.
The (x-2)^2 and (y + 4)^2 parts are the squares. You cannot just add some constants like -4 and -16 to the stuff inside the brackets, because of the ^2.

But there is no need to do that.

See the -4 on both sides of the equation? Add 4 on both sides and poof! the're gone,
leaving us with

(x -2)^2 + (y + 4)^2 - 16 = 0. Now the general form is
(x - a)^2 + (y - b)^2 = r^2.

If we add 16 on both sides we have:

(x -2)^2 + (y + 4)^2 = 16, or:
(x -2)^2 + (y + 4)^2 = 4^2.

Now we can read off the midpoint: it is (2, -4), because a = 2 and b = -4.
Also r = 4.

We're done!

Advice: look up the theory on this, because I have the feeling you do not understand this procedure completely...

Answer 24

Okay i will look up that info thank you for all your help iwould probably still be stuck on this right now. Thanks again

Answer 25

yw!