SOLUTION: Write the equation of the circle that passes through the given points. (7,6),(11,6),(9,4) answer:___________ ___________ ____

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Write the equation of the circle that passes through the given points. (7,6),(11,6),(9,4) answer:___________ ___________ ____      Log On


   

Question 866674: Write the equation of the circle that passes through the given points.
(7,6),(11,6),(9,4)
answer:____________________________
Found 2 solutions by josgarithmetic, richwmiller:
Answer by josgarithmetic(39799) About Me  (Show Source):
You can put this solution on YOUR website!
Using the standard form equation for a circle, you can make three separate equations for a system:

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%287-h%29%5E2%2B%286-k%29%5E2=r%5E2
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%2811-h%29%5E2%2B%286-k%29%5E2=r%5E2
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%289-h%29%5E2%2B%284-k%29%5E2=r%5E2
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Equating the left members of the first two equations, based on both equal to r%5E2 and that both members contain %286-k%29%5E2, allows you to find a value for h.

You can use this value for h to turn the variable expression in each of the three equations into a constant term; so now you will have three equations and just TWO unknowns, k and r.

I started this separately myself on paper, but not yet finished...
This part just for h, but omitting to show the steps of the process: highlight_green%28h=-2%29...

Using that h=-2, the equation Number 2 and Number 3 become:
169%2B%286-k%29%5E2=r%5E2
and
121%2B%284-k%29%5E2=r%5E2
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Now equation each expression for both being r%5E2, this gives 169%2B%286-k%29%5E2=121%2B%284-k%29%5E2.
Performing the arithmetic, algebraic steps gives...
.
highlight_green%28k=17%29.

Use any of the original or transformed equations of the system to compute the value of r.
81%2B%286-%28-17%29%29%5E2=r%5E2
81%2B23%5E2=r%5E2
r%5E2=81%2B529
highlight_green%28r%5E2=610%29-----see that this is still in the squared form.
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The equation of the circle should be: highlight%28%28x%2B2%29%5E2%2B%28y-17%29%5E2=610%29.

Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
Since the circle goes through points
(7,6),(11,6)
the diameter =4 and radius =2
and the center is at (9,6)
(9,4) is also on the circle
(x-9)^2+(y-6)^2=4
These equations are valid and yield the same results as above
(7-h)^2+(6-k)^2=r^2,
(11-h)^2+(6-k)^2=r^2 ,
(9-h)^2+(4-k)^2=r^2
h = 9, k = 6, r = 2
(x-9)^2+(y-6)^2=4