Expert: Sherman D. - 2/21/2008

Question
i am given the endpoints of the diameter of a circle with the points (-2,-3) and (4,-10).  I have to find the standard form of the equation, which represents a circle.  Can you please help me?

Answer
sorry it took so long, for some reason when i clicked the email link on my YM and when it took me to my email, nothing showed up.

since this is the diameter, and you need to put it in (x - h)^2 + (y - k^2) = r^2

find the midpoint of the 2 points

(-2,-3) and (4,-10)

M = (-2 + 4)/2,(-3 + -10)/2
M = (2/2),(-13/2)
M = 1,-6.5

now that we know what the center is, to find the radius, just use the distance formula

(1,-6.5) and (-2,-3)

D = sqrt((-2 - 1)^2 + (-3 - (-6.5))^2)
D = sqrt((-2 + (-1))^2 + (-3 + (13/2))^2)
D = sqrt((-3)^2 + ((-6 + 13)/2)^2)
D = sqrt(9 + (7/2)^2)
D = sqrt(9 + (49/4))
D = sqrt((36 + 49)/4)
D = sqrt(85/4)

now we have

(x - 1)^2 + (y - (-6.5))^2 = sqrt(85/4)^2
(x - 1)^2 + (y + (13/2))^2 = (85/4)
(x - 1)^2 + ((2y + 13)/2)^2 = (85/4)
(x - 1)^2 + (1/4)(2y + 13)^2 = (85/4)
4(x - 1)^2 + (2y + 13)^2 = 85

ANS : 4(x - 1)^2 + (2y + 13)^2 = 85