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How do you find the radius & center of a cricle when given the end points of the diameter?

I know there is a certain formula for this type of problem.
This comes from a college algebra assignment.
the two endpoint (-2,0) and (4,0) are given, it says to find the coordinates of the center & length of the radius, it also says to write an equation of the circle in expanded form?? I think i could find hte answer by graphing, but Id rather know the formula

Suppose (m,n) and (p,q) are the endpoints of a principal diameter. The center is ((m+p)/2,(n+q)/2).
For the circle given, the center is ((-2+4)/2,(0+0)/2) = (1,0).

The radius of a circle is half of the length of a diameter.
d = sqrt[(m-p)^2 + (n-q)^2]
For the circle given, d = sqrt[(-2 - 4)^2 + (0 - 0)^2]
d = sqrt[-6^2]
d = 6
r = d/2
r = 6/2
r = 3

One general equation for a circle is (x-a)^2 + (y-b)^2 = r^2,
where the center of the circle is (a,b).
Given (-2,0) and (4,0) in the circle,
(-2-a)^2 + (0-b)^2 = 3^2 and
(4-a)^2 + (0-b)^2 = 3^2

Subtract these equations.
(-2 - a)^2 - (4-a)^2 = 0
(4 + 4a + a^2) - (16 -8a + a^2) = 0
12a -12 = 0
a = 1

Recall that (4-a)^2 + (0-b)^2 = 3^2.
(4-1)^2 + (0-b)^2 = 3^2
3^2 + (0-b)^2 = 3^2
(0-b)^2 = 0
b = 0

The center of the circle is (1,0).

This circle is (x-1)^2 + y^2 = 3^2.

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admin posted at 2010-1-22 Category: Home Audio

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