SOLUTION: A circle contains the points (0,0), (-7,-17), and (18,-12). Solve a system of three equations in h,k, and r and give the equation of the circle. Show work. (The teacher gave the hi

Question 83412: A circle contains the points (0,0), (-7,-17), and (18,-12). Solve a system of three equations in h,k, and r and give the equation of the circle. Show work. (The teacher gave the hint: Use given (x,y) pairs in to get 3 equations to solve the system using substition or elimination. I tried to do this, but got stuck...can you do both substitution and elimination in the same problem?)
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
A circle contains the points (0,0), (-7,-17), and (18,-12). Solve a system of three equations in h,k, and r and give the equation of the circle. Show work. (The teacher gave the hint: Use given (x,y) pairs in (x-h)^2+(y-k)^2 = r^2 to get 3 equations to solve the system using substitution or elimination.
:
Eq1 (0,0):
(0-h)^2 + (0-k) = r^2
h^2 + k^2 = r^2
:
Eq2 (-7,-17):
(-7-h)^2 + (-17-k)^2 = r^2
(49 + 14h + h^2) + (289 + 34k + k^2) = r^2
h^2 + 14h + k^2 + 34k + 338 = r^2
:
Eq3 (18,-12)
(18-h)^2 + (-12-k)^2 = r^2
324 - 36h + h^2 + 144 + 24k + k^ = r^2
h^2 - 36h + k^2 + 24k + 468 = r^2
:
If you subtract eq 1 from eq 2 you get:
14h + 34k + 338 = 0
or
14h + 34k = -338
:
If you subtract eq 1 from eq 3 you get:
-36h + 24k + 468 = 0
or
-36h + 24k = -468
:
Two equations, two unknowns
:
Multiply the 1st eq by 36 and the 2nd eq by 14 to eliminate h, find k:
504h + 1224k = -12168
-504h + 336k = - 6552
--------------------------adding eliminates h
0h + 1560k = -18720
k = -18720/1560
k = -12
:
Find h using: 14h + 34k = -338; substitute -12 for k
14h + 34(-12) = -338
14h - 408 = - 338
14h = -338 + 408
14h = +70
h = 70/14
h = +5
:
We have h = +5 and k = -12; use h^2 + k^2 = r^2 to find r
:
25 + 144 = r^2
169 = r^
r = sqrt(169)
r = 13
:
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