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

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1. anonymous

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

2. anonymous

can you show me step by step?

3. 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?

4. anonymous

yes sorta.

5. 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?

6. anonymous

yes

7. anonymous

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

8. anonymous

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

9. 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?

10. anonymous

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

11. anonymous

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

12. anonymous

no :/

13. 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...

14. anonymous

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

15. anonymous

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

16. anonymous

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

17. anonymous

x^2+y^2=r?

18. 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$x^2+y^2=r^2$

19. 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?

20. anonymous

ok

21. 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$\left( x-a \right)^2+\left( y-b \right)^2 = r^2$

22. 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?

23. anonymous

4

24. anonymous

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

25. anonymous

yup

26. 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?

27. anonymous

yes a is 2 and b is 4

28. anonymous

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

29. anonymous

4^2?

30. anonymous

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

31. anonymous

What is y - (-4) ?

32. anonymous

b=-4??

33. anonymous

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

34. anonymous

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

35. anonymous

-4x, 8y?

36. 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?

37. anonymous

2,-4?

38. anonymous

39. anonymous

4

40. anonymous

41. anonymous

thanks soooooo muchhh!!!

42. 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!