How do you find the equation of the circle with a diameter that has endpoints at (-7, -2) and (-15, 6)?

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Answer 1 · 2015-12-28T15:19:01

#(x + 11)^2 + (y - 2)^2 = 128#

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

Where #C: (h, k)#

To find the center, use the midpoint formula

#C: ((x_1 + x_2)/2, (y_1 + y_2)/2)#

#=> C: ((-7 + -15)/2, (-2 + 6)/2)#

#=> C: (-11, 2)#

To find the radius, use the distance formula

#r = sqrt((x_1 - x_2)^2 + (y_1 - y_2)^2)#

#=> r = sqrt((-7 - -15)^2 + (-2 - 6)^2)#

#=> r = sqrt(8^2 + (-8)^2)#

#=> r = sqrt(128#

#=> r = 8sqrt2#

Therefore, the equation of the circle is

#(x + 11)^2 + (y - 2)^2 = (8sqrt2)^2#

#=> (x + 11)^2 + (y - 2)^2 = 128#