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Question 583839: Write the standard equation for the circle that passes through the points:
(0, 0)
(6, 0)
(0, - 8)
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Write the standard equation for the circle that passes through the points:
(0, 0)
(6, 0)
(0, - 8
**
Standard form of equation for a circle: (x-h)^2+(y-k)^2=r^2, (h,k) being the (x,y) coordinates of the center, r=radius.
Solving for h, k and r, using given points
..
Equation for point (0,0):
(0-h)^2+(0-k)^2=r^2
eq1) h^2+k^2=r^2
..
Equation for point (6,0):
(6-h)^2+(0-k)^2=r^2
eq2) (6-h)^2+k^2=r^2
..
Equation for point (0,-8):
(0-h)^2+(-8-k)^2=r^2
eq3) h^2+(-8-k)^2=r^2
..
eq1) h^2+k^2=r^2
eq2) (6-h)^2+k^2=r^2
subtract eq2 from eq1 eliminating k^2 and r^2
h^2-(6-h)^2=0
h^2-(36-12h+h^2)=0
h^2-36+12h-h^2=0
12h=36
h=3
..
eq1) h^2+k^2=r^2
eq3) h^2+(-8-k)^2=r^2
subtract eq3 from eq1 eliminating h^2 and r^2
k^2-(-8-k)^2=0
k^2-(64+16k+k^2)=0
k^2-64-16k-k^2=0
-16k=64
k=-4
..
eq1) h^2+k^2=r^2
3^2+(-4)^2=r^2
9+16=r^2
r^2=25
..
Equation of given circle:
(x-3)^2+(y+4)^2=25
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