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Lilmike234

  • one year ago

What is the center and radius of the circle (x+4)^2+(y-2)^2=16

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  1. jim_thompson5910
    • one year ago
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    Hint: Rewrite \[\Large (x+4)^2+(y-2)^2=16\] as \[\Large (x-(-4))^2+(y-2)^2=4^2\] and compare that to the general form\[\Large (x-h)^2+(y-k)^2=r^2\]

  2. lilmike234
    • one year ago
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    So the center is (4,-2) and the radius is 16

  3. UsukiDoll
    • one year ago
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    for the r what is the square root of 16?

  4. lilmike234
    • one year ago
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    4

  5. UsukiDoll
    • one year ago
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    yes... so our radius is actually 4. our center is in the form of (h,k). I think opposites signs are needed. I have to recheck this one fast one.

  6. lilmike234
    • one year ago
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    So the center is (-4,2)

  7. UsukiDoll
    • one year ago
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    ah it's like solving for x and solving for y yeah(-4,2) is the center... radius is 4

  8. jim_thompson5910
    • one year ago
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    \[\Large (x-({\color{red}{-4}}))^2+(y-{\color{blue}{2}})^2={\color{green}{4}}^2\] \[\Large (x-{\color{red}{h}})^2+(y-{\color{blue}{k}})^2={\color{green}{r}}^2\]

  9. UsukiDoll
    • one year ago
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    ok... it's (-4,2) for center radius is 4 ... whew just double checking. I"m 75% sleepy

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