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The equation of a circle is (x - 8)^2 + (y - 5)^2 = 81. Where is (5, 1) located in relation to the circle? A. In the exterior of the circle B. At the center of the circle C. On the circle D. In the interior of the circle

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plug the point into (x-8)^2+(y-5)^2
@myininaya ok i got (5-8)^2+(1-5)^2

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Other answers:

how do i do that?
good what is 1-5?
you mean -4? \[(-3)^2+(-4)^2\]
oh yeah
now what is the square of (-3)?
what is (-3)(-3)?
yes 9 good what is (-4)^2?
In the interior of the circle
ok so what is 16+9
is 25<81?
so it is in the interior
Ohhh ok thanks so much!!! You really helped :)
if we got something > then we would say the exterior if we had got something = to then we would say on the circle
got it! thank you!
and the center of the circle is definitely not (5,1) since it is (8,5)
@myininaya can you help with one more? i'll ask a new question :)
just one more then i must get ready for tonight's event

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