Determine the equation of a circle having a diameter with endpoints at (12, -14) and (2,4)

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Determine the equation of a circle having a diameter with endpoints at (12,-14) and (2,4):

The equation of a circle can be written as `(x-h)^2+(y-k)^2=r^2` where the center is at (h,k) with a radius r.

The center of a circle will be at the midpoint of any diameter. So the center of the circle is at `((12+2)/2,(-14+4)/2)` or `(7,-5)` .

** The midpoint of a segment with endpoints (a,b),(c,d) is `M=((a+c)/2,(b+d)/2)` **

The radius is half of the length of the diameter. We can use the distance formula to find the length of the diameter. The distance between two points `(x_1,y_1),(x_2,y_2)` is `d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

So the length of the diameter is `d=sqrt((12-2)^2+(-14-4)^2)` so `d=sqrt(424)=2sqrt(106)` .

Thus the radius has length `sqrt(106)` and `r^2=106` .

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The center of the circle is at (7,-5) and the square of the radius is 106. The equation is:

`(x-7)^2+(y+5)^2=106`

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