Is A at (3,0) and B at (7,6) ?
if so then the center is halfway between.
Xc = (3+7)/2 = 5
Yc = (0+6)/3 = 3
so of form
(x - 5)^2 + (y - 3)^2 = r^2
but r is half the length of AB
AB^2 = 4^2 + 6^2 = 16+36 = 52
AB = 7.21
(1/2) AB = r = 3.61
r^2 = 13
so in the end
(x - 5)^2 + (y - 3)^2 = 13
Need help please:
(3,0) (7,6)
(x-7)2 + (y-6)2 = 7(2)
(x-7)2 + (y-6)2 = 49
Having difficulty with this problem.Can someone help?
geometry - Reiny, Saturday, March 31, 2012 at 8:07am
You don't say what the actual problem is.
I see the resemblance to finding the equation of a circle??
What are you trying to find?
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geometry - Kel, Saturday, March 31, 2012 at 6:59pm
It says to write an equation of a circle with diameter
__
AB
2 answers
6 = x + 2