Question

**please help!**
Using the following equation, find the center and radius of the circle.
x^2 - 4x + y^2 + 8y = -4

Answer 1

I would complete the square twice, once for a quadratic in x and once for a quadratic in y.

Answer 2

can you show me step by step?

Answer 3

In other words,
Add 4 to the 4^2-4x to create a perfect square trinomial. Then add 16 to the y^2 + 8y to make another perfect square trinomial. Then, because 20 is added to the left hand side, 20 must also be added to th right hand side.\[x^2-4x+y^2+8y=-4\]\[x^2-4x+4+y^2+8y+16=-4+20\]I combined some stuff. Do you understand what I've done?

Answer 4

yes sorta.

Answer 5

Now, consider the LHS as two trinomials. We can group them to make it easier to follow. Plus, we can simplify the RHS\[\left( x^2-4x+4 \right)+\left( y^2+8y+16 \right)=16\]OK with that?

Answer 6

yes

Answer 7

OK. Now, the two trinomials on the LHS are perfect squares. Can you factor them and write them as squares?

Answer 8

would it be (-4^2−4x+4)+(20^2+8y+16)=16

Answer 9

Not quite. Let's slow down a bit and take it step by step. The first trinomial on the LHS is\[x^2-4x+4\]Can you factor this into the product of two binomials?

Answer 10

x^2-4x+4-4+20? im really bad at math and I'm home schooled so it's even tougher

Answer 11

Have you multiplied binomials together before? Stuff like\[\left( x+2 \right)\left( x-3 \right) = x^2-x-6\]

Answer 12

no :/

Answer 13

Oh, then we have a problem. Factoring is the reverse of the above. Let me think of there's another way to get at this...

Answer 14

So you haven't used this technique of 'completing the square' before either?

Answer 15

nope. I took basics of alegebra then was thrown into geometry

Answer 16

OK. Do you know what the equation of a circle looks like?

Answer 17

x^2+y^2=r?

Answer 18

That's very good, but the radius must also squared. For a circle with center at the origin (0, 0), it's equation would be\[x^2+y^2=r^2\]

Answer 19

IN your question, however, the center of the circle is NOT at the origin. So, for a circle with its center at some point (a, b), its equation looks like\[\left( x-a \right)^2 + \left( y-b \right)^2 = r^2\]Notice that, if the center is at the origin, a=b=0 and we get the equation you proposed earlier. Understand?

Answer 20

ok

Answer 21

Alright. So, I've been tying to help you get the equation you were given in the problem\[x^2-4x+y^2+8y=-4\]to look like\[\left( x-a \right)^2+\left( y-b \right)^2 = r^2\]

Answer 22

So far, we've done a couple of things and are at the stage where we have\[\left( x^2-4x+4 \right)+\left( y^2+8y+16 \right)=16\]Look at the RHS. What number squared is equal to 16?

Answer 23

4

Answer 24

Good. So that's the radius. OK with that much?

Answer 25

yup

Answer 26

Good. Now I'm going to cheat a bit and factor those trinomials for you. We have\[\left( x^2-4x+x \right)+\left( y^2+8x+16 \right) = 4^2\]\[\left( x-2 \right)^2+\left( y+4 \right)^2=4^2\]Compare that with\[\left( x-a \right)^2+\left( y-b \right)^2=r^2\]Can you tell what a and b are?

Answer 27

yes a is 2 and b is 4

Answer 28

You're right about a, but look carefully at the signs and try b again.

Answer 29

4^2?

Answer 30

Nope. If b=4, then you'd have (y - 4)^2. But you want (y + 4)^2. What does b have to be?

Answer 31

What is y - (-4) ?

Answer 32

b=-4??

Answer 33

That's right.\[\left( y-\left( -4 \right) \right)^2 = \left( y+4 \right)^2\]So b must be -4.

Answer 34

So now you know a, b, and r. Where's the center of the circle?

Answer 35

-4x, 8y?

Answer 36

No. We know that a=2, b=-4 and r=4. The center of the circle is at coordinates (a, b) and the radius is r. So where;s the center of the circle?

Answer 37

2,-4?

Answer 38

Correct. And what's the radius?

Answer 39

4

Answer 40

There's your answer.

Answer 41

thanks soooooo muchhh!!!

Answer 42

You're welcome. I would encourage you to get some help learning to multiply and factor polynomials. Those skills will be necessary to move forward. Good luck!