anonymous
  • anonymous
**please help!** Using the following equation, find the center and radius of the circle. x^2 - 4x + y^2 + 8y = -4
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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anonymous
  • anonymous
I would complete the square twice, once for a quadratic in x and once for a quadratic in y.
anonymous
  • anonymous
can you show me step by step?
anonymous
  • 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
  • anonymous
yes sorta.
anonymous
  • 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
  • anonymous
yes
anonymous
  • anonymous
OK. Now, the two trinomials on the LHS are perfect squares. Can you factor them and write them as squares?
anonymous
  • anonymous
would it be (-4^2−4x+4)+(20^2+8y+16)=16
anonymous
  • 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
  • anonymous
x^2-4x+4-4+20? im really bad at math and I'm home schooled so it's even tougher
anonymous
  • anonymous
Have you multiplied binomials together before? Stuff like\[\left( x+2 \right)\left( x-3 \right) = x^2-x-6\]
anonymous
  • anonymous
no :/
anonymous
  • 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
  • anonymous
So you haven't used this technique of 'completing the square' before either?
anonymous
  • anonymous
nope. I took basics of alegebra then was thrown into geometry
anonymous
  • anonymous
OK. Do you know what the equation of a circle looks like?
anonymous
  • anonymous
x^2+y^2=r?
anonymous
  • 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
  • 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
  • anonymous
ok
anonymous
  • 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
  • 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
  • anonymous
4
anonymous
  • anonymous
Good. So that's the radius. OK with that much?
anonymous
  • anonymous
yup
anonymous
  • 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
  • anonymous
yes a is 2 and b is 4
anonymous
  • anonymous
You're right about a, but look carefully at the signs and try b again.
anonymous
  • anonymous
4^2?
anonymous
  • 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
  • anonymous
What is y - (-4) ?
anonymous
  • anonymous
b=-4??
anonymous
  • anonymous
That's right.\[\left( y-\left( -4 \right) \right)^2 = \left( y+4 \right)^2\]So b must be -4.
anonymous
  • anonymous
So now you know a, b, and r. Where's the center of the circle?
anonymous
  • anonymous
-4x, 8y?
anonymous
  • 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
  • anonymous
2,-4?
anonymous
  • anonymous
Correct. And what's the radius?
anonymous
  • anonymous
4
anonymous
  • anonymous
There's your answer.
anonymous
  • anonymous
thanks soooooo muchhh!!!
anonymous
  • 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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