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

  2. anonymous
    • one year ago
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    can you show me step by step?

  3. anonymous
    • one year ago
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    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
    • one year ago
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    yes sorta.

  5. anonymous
    • one year ago
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    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
    • one year ago
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    yes

  7. anonymous
    • one year ago
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    OK. Now, the two trinomials on the LHS are perfect squares. Can you factor them and write them as squares?

  8. anonymous
    • one year ago
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    would it be (-4^2−4x+4)+(20^2+8y+16)=16

  9. anonymous
    • one year ago
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    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
    • one year ago
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    x^2-4x+4-4+20? im really bad at math and I'm home schooled so it's even tougher

  11. anonymous
    • one year ago
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    Have you multiplied binomials together before? Stuff like\[\left( x+2 \right)\left( x-3 \right) = x^2-x-6\]

  12. anonymous
    • one year ago
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    no :/

  13. anonymous
    • one year ago
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    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
    • one year ago
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    So you haven't used this technique of 'completing the square' before either?

  15. anonymous
    • one year ago
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    nope. I took basics of alegebra then was thrown into geometry

  16. anonymous
    • one year ago
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    OK. Do you know what the equation of a circle looks like?

  17. anonymous
    • one year ago
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    x^2+y^2=r?

  18. anonymous
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    ok

  21. anonymous
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    4

  24. anonymous
    • one year ago
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    Good. So that's the radius. OK with that much?

  25. anonymous
    • one year ago
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    yup

  26. anonymous
    • one year ago
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    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
    • one year ago
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    yes a is 2 and b is 4

  28. anonymous
    • one year ago
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    You're right about a, but look carefully at the signs and try b again.

  29. anonymous
    • one year ago
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    4^2?

  30. anonymous
    • one year ago
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    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
    • one year ago
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    What is y - (-4) ?

  32. anonymous
    • one year ago
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    b=-4??

  33. anonymous
    • one year ago
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    That's right.\[\left( y-\left( -4 \right) \right)^2 = \left( y+4 \right)^2\]So b must be -4.

  34. anonymous
    • one year ago
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    So now you know a, b, and r. Where's the center of the circle?

  35. anonymous
    • one year ago
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    -4x, 8y?

  36. anonymous
    • one year ago
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    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
    • one year ago
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    2,-4?

  38. anonymous
    • one year ago
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    Correct. And what's the radius?

  39. anonymous
    • one year ago
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    4

  40. anonymous
    • one year ago
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    There's your answer.

  41. anonymous
    • one year ago
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    thanks soooooo muchhh!!!

  42. anonymous
    • one year ago
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    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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