SOLUTION: Find the center and the radius of the circle that passes through the points (4,4) , (1,3) and (8,-4). What are the coordinates of the center? Thank you in advanced.

Algebra ->  Linear-equations -> SOLUTION: Find the center and the radius of the circle that passes through the points (4,4) , (1,3) and (8,-4). What are the coordinates of the center? Thank you in advanced.      Log On


   

Question 985229: Find the center and the radius of the circle that passes through the points (4,4) , (1,3) and (8,-4).
What are the coordinates of the center?
Thank you in advanced.
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
Find the center and the radius of the circle that passes through the points
(4,4) , (1,3) and (8,-4).

equation of the circle is:
%28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2 where h and k are x and y coordinates of the center, and r is radius
so use given points to set up system of three unknown
(4,4)
%284-h%29%5E2%2B%284-k%29%5E2=r%5E2....simplify
16-8h%2Bh%5E2%2B16-8k%2Bk%5E2=r%5E2.............eq.1
(1,3)
%281-h%29%5E2%2B%283-k%29%5E2=r%5E2
1-2h%2Bh%5E2%2B9-6k%2Bk%5E2=r%5E2.............eq.2

(8,-4)
%288-h%29%5E2%2B%28-4-k%29%5E2=r%5E2....simplify
64-16h%2Bh%5E2%2B16%2B8k%2Bk%5E2=r%5E2.............eq.3
start with
16-8h%2Bh%5E2%2B16-8k%2Bk%5E2=r%5E2.............eq.1
64-16h%2Bh%5E2%2B16%2B8k%2Bk%5E2=r%5E2.............eq.3
--------------------------------------------------------------------subtract
16-8h%2Bh%5E2%2B16-8k%2Bk%5E2-%2864-16h%2Bh%5E2%2B16%2B8k%2Bk%5E2%29=r%5E2-r%5E2

-8h%2B16-8k-64%2B16h-8k=0
8h-16k=64-16.......both sides divide by 8
h-2k=8-2
h-2k=6...............eg.1a

go with
16-8h%2Bh%5E2%2B16-8k%2Bk%5E2=r%5E2.............eq.1
1-2h%2Bh%5E2%2B9-6k%2Bk%5E2=r%5E2.............eq.2
---------------------------------------------------------subtract
16-8h%2Bh%5E2%2B16-8k%2Bk%5E2-%281-2h%2Bh%5E2%2B9-6k%2Bk%5E2%29=r%5E2-r%5E2

22-8h-8k%2B2h%2B6k=0

-6h-2k=-22...............eg.2a
now go with
h-2k=6...............eg.1a
-6h-2k=-22...............eg.2a
-----------------------------------------------subtract
h-2k-%28-6h-2k%29=6-%28-22%29
h-cross%282k%29%2B6h%2Bcross%282k%29=6%2B22
h%2B6h=28
7h=28
h=28%2F7
highlight%28h=4%29
go to
h-2k=6...............eg.1a...plug in 4 for h
4-2k=6
4-6=2k
-2=2k
k=+-2%2F2
highlight%28k=+-1%29
go to
16-8h%2Bh%5E2%2B16-8k%2Bk%5E2=r%5E2.............eq.1 ..plug in 4 for h and -1 for k and find r
16-8%284%29%2B4%5E2%2B16-8%28-1%29%2B%28-1%29%5E2=r%5E2
cross%2816%29-cross%2832%29%2Bcross%2816%29%2B16%2B8%2B1=r%5E2
16%2B8%2B1=r%5E2
25=r%5E2
highlight%28r=5%29
so, your equation is: %28x-4%29%5E2%2B%28y%2B1%29%5E2=25

check: