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Mathematics 8 Online
OpenStudy (anonymous):

@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.

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

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)

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

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

OpenStudy (anonymous):

25=5

jimthompson5910 (jim_thompson5910):

so the radius is 5

jimthompson5910 (jim_thompson5910):

r = 5 r^2 = 25

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