Question

using the following equation, find the center and radius of the circle. x^2 + 4x + y^2 - 6y= -4

Answer 1

Please help I have none of my formulas!

Answer 2

you have to complete the squares to get the equation of a circle in standard form

Answer 3

What? How can I complete the squares when it's y and x

Answer 4

you have to complete the square with x and y, then you wil get:
(x-a)^2 + (y-b)^2 = r^2
x-a is the change in x
y-b is the change in y
r is the radius

Answer 5

I'm sorry I'm just trying to understand that. So how to I get the radius, r, if that's all letters? I must be being very thick right now I'm sorry

Answer 6

.?

Answer 7

Okay, you've got

x^2 + 4x + y^2 - 6y = -4

you need to get (x-a)^2 + (y-b)^2 = r^2

x^2 + 4x + <some number> is a perfect square if you can find a number (we'll call it c)

such that

(x+c)^2 = (x+c)(x+c) = x^2 + 4x + <some number>

Answer 8

(x+c)(x+c) = x^2 + cx + cx + c^2
= x^2 + 2cx + c^2

if we can find a value of c so that

4x = 2cx, then we can write x^2 + 4x + <something> = (x+c)^2

and that will be completing the square on x. then we'll do the same procedure to complete the square on y.

Answer 9

I seriously have no idea. Do I have to find the number that x^2+ 4x equals?

Answer 10

4x = 2cx

can you solve that for c?

Answer 11

hint: you can divide both sides by x, to get

4x/x = 2cx/x
or
4 = 2c

Answer 12

Then
4=2c
/2. /2
2=c

Answer 13

right!

so (x+c)^2 = (x+2)^2 = (x+2)(x+2) = x^2 + 2x + 2x + 4 = x^2 + 4x + 4

well, that's just like the x stuff we have in our original equation, right? we just need to find a spare "+4" somewhere :-)

Answer 14

we had

x^2 + 4x + y^2 - 6y= -4

we'd like to rewrite the x^2 + 4x part as (x+c)^2 = (x+2)^2

but to do that we need a "+4" on the left side of the equation. we can just make one!

4 - 4 = 0, right?

so we can insert

x^2 + 4x + 4 - 4 + y^2 - 6y = -4

we haven't changed the value of the left side or the right side by doing that.

[x^2 + 4x + 4] - 4 + y^2 - 6y = -4

now we can rewrite the stuff in the brackets as (x+2)^2

(x+2)^2 - 4 + y^2 - 6y = -4

progress! now we need to do the same process with y^2 - 6y

Answer 15

does that make sense so far?

Answer 16

if y^2 - 6y + <something> = y^2 + 2cy + c^2

-6y = 2cy
-6 = 2c
c = -3

so we can rewrite y^2 - 6y + 9 as

(y-3)^2

Answer 17

we just need to find (or create) a +9 in our equation.

(x+2)^2 - 4 + y^2 - 6y = -4

(x+2)^2 - 4 + y^2 - 6y + 9 - 9 = -4
(x+2)^2 -4 + [y^2 - 6y + 9] - 9 = -4
replace the stuff in brackets by (y-3)^2
(x+2)^2 - 4 + (y-3)^2 - 9 = -4

now we just move the remaining numbers to the right hand side
(x+2)^2 -4 + 4 + (y-3)^2 - 9 + 9 = - 4 + 4 + 9
(x+2)^2 + (y-3)^2 = 9

that's the form we want:

(x-h)^2 + (y-k)^2 = r^2

to make those match, we have

(x+2) = (x-h)
2 = -h
h = -2

(y-3) = (y-k)
-3 = -k
k = 3

r^2 = 9
r = 3

Answer 18

the general form of the circle equation:

(x-h)^2 + (y-k)^2 = r^2

gives us the center at (h,k) and the radius is r

if our equation is (x+2)^2 + (y-3)^2 = 9 and
h = -2, k = 3, and r = 3

what is the radius? what is the center?

Answer 19

radius is 3, because r = 3
center is at (-2, 3) because center is at (h, k) and h = -2, k = 3

Answer 20

You are very nice thanks for explaining that all! Honestly I didn't study at all because my end date is today and I have to finish this so I have no idea what I'm doing so thank you!

Answer 21

I offer a full money-back guarantee. If tomorrow this still doesn't make sense, tag me on a question and I'll try explaining it again until it does :-)

Answer 22

Haha ok thank you