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# What is the standard form of the equation of a circle with endpoints of a diameter at the points (7,8) and (-5,6)?

the standard form formula is ##(x-h)^2 + (y-k)^2 = r^2 ##

therefore ..

##(x-1)^2 + (y-7)^2 = (2sqrt37)^2 ## ##(x-1)^2 + (y-7)^2 = 148 ##

to be able to find the of a circle we need to find the value of radius and the point where the center lies..

to find the center .. we need to use the midpoint of the diameter using the midpoint formula..

##m = ((x_1+x_2)/2 , (y_1 + y_2 )/2)## ##m = ((7+(-5))/2 , (8 + 6 )/2)## ##m = (1 , 7)##

to find the radius of the circle we need to use the distance formula

## d = sqrt((x_2-x_1)^2+(y_2-y_1)^2## ## d = sqrt((-5-7)^2+(6-8)^2## ## d = sqrt((-12)^2+(-2)^2## ## d = sqrt(144+4## ## d = 2sqrt37##

the standard form formula is ##(x-h)^2 + (y-k)^2 = r^2 ##

therefore ..

##(x-1)^2 + (y-7)^2 = (2sqrt37)^2 ## ##(x-1)^2 + (y-7)^2 = 148 ##