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anonymous

  • one year ago

What is the standard form of the equation of a circle with its center at (2, -3) and passing through the point (-2, 0)? (x − 2)squared + (y + 3)squared = 5 (x + 2)squared + (y − 3)squared = 25 (x − 2)squared + (y + 3)squared = 25 (x − 2)squared − (y + 3)squared = 5

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  1. anonymous
    • one year ago
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    @Deezzz

  2. anonymous
    • one year ago
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    @welshfella

  3. anonymous
    • one year ago
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    @rvc

  4. welshfella
    • one year ago
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    the general form is (x - a)^" + (y - b)^2 = r^2 where (a,b) is the center and r is the radius

  5. anonymous
    • one year ago
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    yes

  6. welshfella
    • one year ago
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    so you should be able to work out the left side now by plugging in a = 2 and b = -3

  7. welshfella
    • one year ago
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    and the radius will be the distance between the center and the point (-2,0)

  8. anonymous
    • one year ago
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    thanks

  9. welshfella
    • one year ago
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    yw - if you need more help let me know

  10. rvc
    • one year ago
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    :) all the best

  11. anonymous
    • one year ago
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    @Deezzz

  12. anonymous
    • one year ago
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    @nincompoop

  13. anonymous
    • one year ago
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    @Hero

  14. anonymous
    • one year ago
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    @triciaal

  15. Hero
    • one year ago
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    In general, the standard form of the equation of a circle is \((x - h)^2 + (y - k)^2 = r^2\). In this case, (h,k) = (2,3) and (x,y) = (-2,0). Plug both points in to the equation, then simplify and solve for r. Next re-write the standard form of the equation, except this time, insert the given center point and the value of r.

  16. anonymous
    • one year ago
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    ok ill try, thanks

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