What is the standard form of the equation of a circle with endpoints of the diameter at (0,10) and (-10,-2)?

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Answer 1 · 2016-03-24T16:35:05

#(x + 5)^2 + (y - 4)^2 = 61#

The equation of a circle in standard form is

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

where
#h#: #x#-coordinate of the center
#k#: #y#-coordinate of the center
#r#: radius of the circle

To get the center, get the midpoint of the endpoints of the diameter

#h = (x_1 + x_2)/2

#=> h = (0 + -10)/2#
#=> h = -5#

#k = (y_1 + y_2)/2#

#=> k = (10 + -2)/2#
#=> k = 4#

#c: (-5, 4)#

To get the radius, get the distance between the center and either endpoint of the diameter

#r = sqrt((x_1 - h)^2 + (y_1 - k)^2)#

#r = sqrt((0 - -5)^2 + (10 - 4)^2)#

#r = sqrt(5^2 + 6^2)#

#r = sqrt61#

Hence, the equation of the circle is

#(x - -5)^2 + (y - 4)^2 = (sqrt61)^2#

#=> (x + 5)^2 + (y - 4)^2 = 61#