SOLUTION: I used the formula (x-h)^2 +(y-k)^2=r^2 to determine the equation of the circle. I got c=(4,0)r=sqrt 13 (x-4)^2+(y-0)^2=(sqrt13)^2 (x-4)^2=(y-0)^2 =13 but now I have to determine

Algebra ->  Circles -> SOLUTION: I used the formula (x-h)^2 +(y-k)^2=r^2 to determine the equation of the circle. I got c=(4,0)r=sqrt 13 (x-4)^2+(y-0)^2=(sqrt13)^2 (x-4)^2=(y-0)^2 =13 but now I have to determine       Log On


   

Question 308795: I used the formula (x-h)^2 +(y-k)^2=r^2 to determine the equation of the circle. I got c=(4,0)r=sqrt 13 (x-4)^2+(y-0)^2=(sqrt13)^2
(x-4)^2=(y-0)^2 =13 but now I have to determine where the circle intersects the x-axis using a system of equations. Can you please show me how to do this.
Found 2 solutions by stanbon, Fombitz:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
(x-4)^2+(y-0)^2 =13 but now I have to determine where the circle intersects the x-axis using a system of equations.
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Let y = 0; then solve for "x":
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(x-4)^2 = 13
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x-4 = +-sqrt(13)
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x = 4+sqrt(13) or x = 4-sqrt(13)
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Cheers,
Stan H.

Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
When you're on the x-axis, then y=0.
%28x-4%29%5E2%2B%28y-0%29%5E2+=13
%28x-4%29%5E2%2B%280-0%29%5E2+=13
%28x-4%29%5E2=13
I think you can take it from there to solve for the two x values.
Remember that then y=0 in the ordered pair.