@jim_thompson5910 Challenge: Graph the equation with a diameter that has endpoints at (3, -4) and (-5, 2). Label the center and at least four points
on the circle. Write the equation of the circle.
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jimthompson5910 (jim_thompson5910):
Now we don't have the center, but we can find it
jimthompson5910 (jim_thompson5910):
We just need to find the midpoint of the segment with endpoints of (3, -4) and (-5, 2)
OpenStudy (anonymous):
can you set it :) @jim_thompson5910
jimthompson5910 (jim_thompson5910):
you average the corresponding coordinates
jimthompson5910 (jim_thompson5910):
so for the x, you add them up and divide by 2
3 + (-5) = -2
-2/2 = -1
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jimthompson5910 (jim_thompson5910):
The x coordinate of the midpoint is x = -1
jimthompson5910 (jim_thompson5910):
how about the y coordinate?
OpenStudy (anonymous):
-2? @jim_thompson5910
jimthompson5910 (jim_thompson5910):
close, but you would divide that by 2 to get -1
jimthompson5910 (jim_thompson5910):
so the midpoint is the point (-1,-1)
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jimthompson5910 (jim_thompson5910):
that's your center
OpenStudy (anonymous):
whats midpoint?
OpenStudy (anonymous):
oh ok
OpenStudy (anonymous):
and the radius is what?
jimthompson5910 (jim_thompson5910):
the distance from the center to either point
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OpenStudy (anonymous):
so how many shoul i go likee what do you suggest ? @jim_thompson5910
jimthompson5910 (jim_thompson5910):
use the distance formula again
jimthompson5910 (jim_thompson5910):
so let's say we want to find the distance from the center (-1,-1) to (3, -4), we would do...
d = sqrt((x2-x1)^2+(y2-y1)^2)
d = sqrt((3-(-1))^2+( -4-(-1))^2)
d = sqrt((3+1)^2+( -4+1)^2)
d = sqrt((4)^2+(-3)^2)
d = ??
OpenStudy (anonymous):
16 +9
jimthompson5910 (jim_thompson5910):
keep going
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