Part of an Ancient wheel....circles?

Question

Three points on the rim of hte wheel part are locatd at A(-3,-3),B(-1,11) and C(5,13).What is the conclusionas to the length of the radius of the original wheel?

Answer 1

Use the equation of a circle.
(x-a)^2+(y-b)^2=r^2

Plug your 3 points in this equation to get 3 new equations.
The point (-3,-3) gives:
(-3-a)^2 + (-3-b)^2=r^2
9+6a+a^2 + 9+6b+b^2 = r^2
18 + 6a + a^2 + 6b + b^2 = r^2

The point (-1,11) gives:
(-1-a)^2+(11-b)^2= r^2
1+2a+a^2 + 121-22b+b^2 = r^2
122 + 2a + a^2 -22b + b^2 = r^2

The point (5, 13) gives:
(5-a)^2+(13-b)^2 = r^2
25 -10a +a^2 + 169-26b+b^2 = r^2
194 - 10a + a^2 -26b +b^2 = r^2

Now we have 3 equations and 3 unknowns.
subtract the second equation from the first.
18 + 6a + a^2 + 6b + b^2 = r^2
122 + 2a + a^2 -22b + b^2 = r^2
--------
-104 + 4a +28b = 0
4a + 28b = 104

Subtract the third equation from the second
122 + 2a + a^2 -22b + b^2 = r^2
194 - 10a + a^2 -26b +b^2 = r^2
----------
-72 +12a + 4b = 0
12a + 4b = 72

multiply both sides of 4a + 28b = 104 by 3
12a + 84b = 312
Subtract 12a + 4b = 72
-------------
80b = 240
b = 3

Substitute in 12a + 4b = 72
12a + 4*3 = 72
12a + 12 = 72
a + 1 = 6
a = 5

Substitute a and b back into the original equation along with the point (-3, -3)
(-3 - 5)^2+(-3-3)^2 = r^2
(-8)^2+(-6)^2 = r^2
64+36 = r^2
r^2 = 100
r = 10
This is your radius.
The equation for the circle is (x - 5)^2+(x-3)^2= 10^2
It's a good idea to try all 3 points in this equation to check your work.

Answer 2

The radius of that circle would be 10. The answer can be reached by intersecting the perpendicular bisectors of the 2 chords formed by these three points. You'll have to determine what formulae to use from your basic trigonometry.