briana.img
  • briana.img
Graphing a circle. How to figure out the correct integers for the points?!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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briana.img
  • briana.img
I always thought the integers were the same
briana.img
  • briana.img
I don't understand how they're positive I thought they would be negative
myininaya
  • myininaya
what are you talking about exactly?

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briana.img
  • briana.img
@myininaya I thought the center coordinates for the second would be (-2,-1), but they're positive.. why is that?
myininaya
  • myininaya
x-2=0 when x=2 and y-1=0 when y=1 so the center is (2,1)
myininaya
  • myininaya
you can do that if confused to compare the equation to \[(x-h)^2+(y-k)^2=r^2 \text{ where } (h,k) \text{ is center }\]
myininaya
  • myininaya
\[(x-h)^2+(y-k)^2=r^2 \\ (x-2)^2+(y-1)^2=(\sqrt{18})^2\]
briana.img
  • briana.img
@myininaya so just set it equal to zero to get the coordinates?
myininaya
  • myininaya
though you should see when comparing these that 2 is in the place of h and 1 is in the place of k
myininaya
  • myininaya
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: \[(x-h)^2+(y-k)^2=r^2 \]
briana.img
  • briana.img
@myininaya so basically just it equal to zero
myininaya
  • myininaya
oh and that was a 16 not a 18 i have bad eyes
myininaya
  • myininaya
yes if you have it in the form: \[(x-h)^2+(y-k)^2=r^2\] then yes
briana.img
  • briana.img
@myininaya ok thx!!!

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