If the endpoints of the diameter of a circle are (−8, 0) and (−12, 2), what is the standard form equation of the circle?

Answer:The standard form equation of circle is[tex](x+10)^{2}[/tex] +[tex](y-1)^{2}[/tex] = 5Step-by-step explanation:Given the endpoints of diameter of circle (-8,0) and (-12,2)The equation of circle is given by [tex](x-h)^{2}[/tex] +[tex](y-k)^{2}[/tex] = [tex]r^{2}[/tex]where (x,y) is any point on circle, (h,k) is center of circle and r is radius of circleTo find (h, k): center of circle is given by the midpoint of diameterMidpoint of diameter with endpoints (x1,y1) and (x2,y2) is(  [tex]\frac{x1+x2}{2}[/tex] , [tex]\frac{y1+y2}{2}[/tex]  )Substituting the values of endpoints(  [tex]\frac{-8-12}{2}[/tex] , [tex]\frac{0+2}{2}[/tex] )(-10,1)(h,k) is (-10,1)Substituting the values of (h,k) and (x,y) as (-10,1) and (-8,0) respectively in equation of  circle we get,[tex](-8+10)^{2}[/tex] + [tex](0-1)^{2}[/tex] = [tex]r^{2}[/tex][tex]r^{2}[/tex] = 5Now substituting the values  of (h,k) and [tex]r^{2}[/tex] , we get The standard form equation of circle as[tex](x+10)^{2}[/tex] +[tex](y-1)^{2}[/tex] = 5