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Do you know the standard equation for a circle?

x^2+y^2=r^2 ??

\[(x-h)^2+(y-k)^2 = r^2\]

Where (h, k) are the center point

ok with (h,k) being the center and r is the radius ..what nxt..

Pick a point, (7,2):
It can be odd sometimes, because the values may change sign

(x-4)^2+(y+2)^2=r^2

ok
so (3)^2+(4)^2=r^2
9+16=r^2
25=r^2
5=r
?

So check if the points are in the circle

idk how to...??

which point??

Hmm, I think I did something wrong - (7,2) is in the circle, but not (1, -6)

but we have to show that those points are on the circle

hold on

ok

sorry, I can't help you. It's been too long

yes

hope it makes sense

so \[\sqrt{(7-1)^2+(2+6)^2}=10\]

oh \[\sqrt{(4-1)^2+(-2+6)^2}=5\]

and \[\sqrt{(4-7)^2+(-2-2)^2}=5\]

that works.... so you have shown the points are the same distance from the fixed point

cool thanks