Question 449407
Write the equation of a circle in standard form if the endpoints of the diameter are (-8,2) and (6,-4)
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Use distance formula to calculate length of diameter:
Diameter=sqrt(x2-x1)^2+(y2-y1)^2
  =sqrt(-8-6)^2+(2-(-4))^2
  =sqrt(-14^2+6^2)
  =sqrt(232)
 Diameter=15.23
 Radius=7.61
 Radius^2=57.91
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Use midpoint formula to find center of circle:
(x,y)=(x1+x2)/2,(y1+y2)/2
       =(-8+6)/2,(2-4)/2
       =-2/2,-2/2
(x,y)=(-1,-1) (center of circle)
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Standard form for equation of circle:(x-h)^2+(y-k)^2=r^2, (h,k)= (x,y) coordinates of center and r= radius
For equation of given circle:
(x+1)^2+(y+1)^2=57.91
see graph of circle below:
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y=(57.91-(x+1)^2)^.5-1
{{{ graph( 300, 300, -10, 10, -10, 10, (57.91-(x+1)^2)^.5-1,-(57.91-(x+1)^2)^.5-1) }}}