SOLUTION: Find the equation of the circle that passes through (8,3) and (2,3) and has its centre on the x-axis. Thank-you!

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Question 917283: Find the equation of the circle that passes through (8,3) and (2,3) and has its centre on the x-axis.
Thank-you!
Found 2 solutions by josgarithmetic, lwsshak3:
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
The center on the x-axis and those two given points form an isosceles triangle, and the x-coordinate of the center of the circle should be in the middle of x=8 and x=3.

That center has y=0, because the point is given as ON THE x-AXIS.

Center is the point (5,0).

Use either of the given points and Distance Formula to calculate and then compute the radius of the circle.

Now just fill-in the necessary values for the standard form equation of a circle.

sqrt%28%288-5%29%5E2%2B%283-0%29%5E2%29
sqrt%28%283%29%5E2%2B%283%29%5E2%29
sqrt%2818%29

When you square that, you have 18, which is r%5E2 for the standard form equation.

highlight%28%28x-5%29%5E2%2By%5E2=18%29

Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
Find the equation of the circle that passes through (8,3) and (2,3) and has its centre on the x-axis.
***
draw a line connecting the 2 given points on the circle.
This line is a chord of the circle
Draw a perpendicular line from the midpoint of the chord to the x-axis
The point at which this line meets the x-axis is the center of the circle.
..
midpoint of the chord=(8+2)/2=10/2=5
center of circle=(5,0)=(h,k)
Form of equation for a circle: (x-h)^2+(y-k)^2=r^2
Using one of given points(2,3) to find radius^2
(2-5)^2+(3-0)^2=r^2
3^2+3^2=9+9=18
r^2=18
equation for circle: (x-5)^2+y^2=18