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anonymous

  • one year ago

Please help!!!! A circle is centered at the point (5, -4) and passes through the point (-3, 2). The equation of this circle in standard form is

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  1. freckles
    • one year ago
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    ok first state the standard form of a circle

  2. anonymous
    • one year ago
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    (x-h)^2 + (y-k)^2=r^2

  3. freckles
    • one year ago
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    great

  4. freckles
    • one year ago
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    now (h,k) is the center of the circle

  5. freckles
    • one year ago
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    you have these values h and k are given

  6. freckles
    • one year ago
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    your equation is so far: (x-5)^2+(y+4)^2=r^2 now you are also given a point (x,y) that is on the circle this will allow you to solve for r

  7. freckles
    • one year ago
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    That point being (-3,2) so replace x with -3 and y with 2 and solve for r

  8. freckles
    • one year ago
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    Please let know what you get for r.

  9. anonymous
    • one year ago
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    9^2?

  10. freckles
    • one year ago
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    let me check

  11. freckles
    • one year ago
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    \[(-3-5)^2+(2+4)^2=r^2 \\ (-8)^2+(6)^2=r^2\] did you get this far?

  12. anonymous
    • one year ago
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    yes

  13. freckles
    • one year ago
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    and so you winded up with 64+36=r^2 right?

  14. anonymous
    • one year ago
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    yeah and i got confused

  15. freckles
    • one year ago
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    well 64+36 should be 100

  16. freckles
    • one year ago
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    you are really done because the standard form of a circle is (x-h)^2+(y-k)^2=r^2 where we needed values h,k , and r^2 we just found r^2 and we were given h and k

  17. anonymous
    • one year ago
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    thank you soo much

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