I would complete the square twice, once for a quadratic in x and once for a quadratic in y.
can you show me step by step?
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?
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?
OK. Now, the two trinomials on the LHS are perfect squares. Can you factor them and write them as squares?
would it be (-4^2−4x+4)+(20^2+8y+16)=16
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?
x^2-4x+4-4+20? im really bad at math and I'm home schooled so it's even tougher
Have you multiplied binomials together before? Stuff like\[\left( x+2 \right)\left( x-3 \right) = x^2-x-6\]
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...
So you haven't used this technique of 'completing the square' before either?
nope. I took basics of alegebra then was thrown into geometry
OK. Do you know what the equation of a circle looks like?
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\[x^2+y^2=r^2\]
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?
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\[\left( x-a \right)^2+\left( y-b \right)^2 = r^2\]
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?
Good. So that's the radius. OK with that much?
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?
yes a is 2 and b is 4
You're right about a, but look carefully at the signs and try b again.
Nope. If b=4, then you'd have (y - 4)^2. But you want (y + 4)^2. What does b have to be?
What is y - (-4) ?
That's right.\[\left( y-\left( -4 \right) \right)^2 = \left( y+4 \right)^2\]So b must be -4.
So now you know a, b, and r. Where's the center of the circle?
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?
Correct. And what's the radius?
There's your answer.
thanks soooooo muchhh!!!
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!