How do you write an equation of a circle whose diameter of the circle are (-2,6) and (-8,10)?
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Please see the explanation.
#(x - -5)^2 + (y - 8)^2 = (sqrt(13))^2#
The general equation of a circle is:
#(x - h)^2 + (y - k)^2 = r^2" [1]"#
where #(x, y)# is any point on the circle, #(h,k)# is the center, and r is the radius.
To go from #(-2, 6) " to " (-8, 10)# one must move to the left 6 and up 4, therefore, the center must be to the left 3 and up 2; the point #(-5, 8)#. Substitute the center into equation [1]:
#(x - -5)^2 + (y - 8)^2 = r^2" [2]"#
Substitute the point #(-2,6)# into equation [2]:
#(-2 - -5)^2 + (6 - 8)^2 = r^2#
Solve for r:
#(3)^2 + (-2)^2 = r^2#
#13 = r^2#
#r = sqrt(13)#
Substitute #r = sqrt(13)# into equation [2]:
#(x - -5)^2 + (y - 8)^2 = (sqrt(13))^2#
This is the equation of the circle.
