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

- anonymous

- jamiebookeater

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- anonymous

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

- anonymous

can you show me step by step?

- anonymous

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?

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- anonymous

yes sorta.

- anonymous

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?

- anonymous

yes

- anonymous

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

- anonymous

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

- anonymous

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?

- anonymous

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

- anonymous

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

- anonymous

no :/

- anonymous

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...

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

x^2+y^2=r?

- anonymous

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\[x^2+y^2=r^2\]

- anonymous

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?

- anonymous

ok

- anonymous

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\[\left( x-a \right)^2+\left( y-b \right)^2 = r^2\]

- anonymous

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?

- anonymous

4

- anonymous

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

- anonymous

yup

- anonymous

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?

- anonymous

yes a is 2 and b is 4

- anonymous

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

- anonymous

4^2?

- anonymous

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

- anonymous

What is y - (-4) ?

- anonymous

b=-4??

- anonymous

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

- anonymous

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

- anonymous

-4x, 8y?

- anonymous

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?

- anonymous

2,-4?

- anonymous

Correct. And what's the radius?

- anonymous

4

- anonymous

There's your answer.

- anonymous

thanks soooooo muchhh!!!

- anonymous

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!

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