How do you write an equation for for circle given that the endpoints of the diameter are (-2,7) and (4,-8)?

`"given the endpoints of the diameter then the centre is at"`
`"the midpoint and the radius is the distance from the "`
`"centre to either of the 2 endpoints"`

`"the equation of a circle in standard form is"`

`color(red)(bar(ul(|color(white)(2/2)color(black)((x-a)^2+(y-b)^2=r^2)color(white)(2/2)|)))`

`"where "(a,b)" are the coordinates of the centre and r "`
`"is the radius"`

`"midpoint "=[1/2(-2+4),1/2(7-8)]=(1,-1/2)`

`"to calculate the radius use the "color(blue)"distance formula"`

`•color(white)(x)d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

`"let "(x_1,y-1)=(4,-8)" and "(x_2,y_2)=(1,-1/2)`

`d=sqrt((1-4)^2+(-1/2+8)^2)`

`color(white)(d)=sqrt(9+225/4)=sqrt261/2`

`(x-1)^2+(y+1/2)^2=(sqrt261/2)^2`

`rArr(x-1)^2+(y+1/2)^2=261/4larrcolor(blue)"equation of circle"`

Center coordinates `O ((4-2)/2, (-7+8)/2) = O(1, -(1/2)`

diameter /2 = radius = `r = sqrt((4+2)^2 + (-8-7)^2) /2 = color(blue)(8.0777)`

Standard equation of a circle is

`(x - h)^2 + (y - k)^2 = r^2`

Therefore, equation of the given circle is

`color(red)((x - 1)^2 + (y + (1/2))^2 = (8.0777)^2)`