need explination: find equation of a circle with in standard form with a diameter tht has endpoints at 6,-2 and -4,4

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need explination: find equation of a circle with in standard form with a diameter tht has endpoints at 6,-2 and -4,4

Mathematics
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no body knows how to do this?
I know wait a second :D
|dw:1328379995392:dw|

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so equation of the circle is (x-1)^2+(y-1)^2=136
The equation you have doesnt satisfy the points though. I think this is the process: First we find the length of the diameter, we just make a triangle from the two points|dw:1328380595002:dw| Using the pythagorean thereom, we can find the length of the diameter which is root 136 But we know that the radius is half of that, which is root 136 /2. Finding the centre of the diameter, as nena has done already, the centre of the circle is (1,1) So plugging all this in, we get (x-1)^2 + (y-1)^2 = 34 The 34 comes from r^2, which is root(136)/2 all squared = 136/4 = 34

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