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briana.img

  • one year ago

Graphing a circle. How to figure out the correct integers for the points?!

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  1. briana.img
    • one year ago
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    I always thought the integers were the same

  2. briana.img
    • one year ago
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    I don't understand how they're positive I thought they would be negative

  3. myininaya
    • one year ago
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    what are you talking about exactly?

  4. briana.img
    • one year ago
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    @myininaya I thought the center coordinates for the second would be (-2,-1), but they're positive.. why is that?

  5. myininaya
    • one year ago
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    x-2=0 when x=2 and y-1=0 when y=1 so the center is (2,1)

  6. myininaya
    • one year ago
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    you can do that if confused to compare the equation to \[(x-h)^2+(y-k)^2=r^2 \text{ where } (h,k) \text{ is center }\]

  7. myininaya
    • one year ago
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    \[(x-h)^2+(y-k)^2=r^2 \\ (x-2)^2+(y-1)^2=(\sqrt{18})^2\]

  8. briana.img
    • one year ago
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    @myininaya so just set it equal to zero to get the coordinates?

  9. myininaya
    • one year ago
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    though you should see when comparing these that 2 is in the place of h and 1 is in the place of k

  10. myininaya
    • one year ago
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    yeah if it confused you to compare it to the equation above but you still need to make sure you have it in the form: \[(x-h)^2+(y-k)^2=r^2 \]

  11. briana.img
    • one year ago
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    @myininaya so basically just it equal to zero

  12. myininaya
    • one year ago
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    oh and that was a 16 not a 18 i have bad eyes

  13. myininaya
    • one year ago
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    yes if you have it in the form: \[(x-h)^2+(y-k)^2=r^2\] then yes

  14. briana.img
    • one year ago
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    @myininaya ok thx!!!

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