Question 672291: A circle cuts the x-axis at (4,0) and (14,0), and cuts the y-axis at (0,6) and (0,8). Find its equation.
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
If the circle cuts the  -axis at ) and ) , then those two points are the endpoints of a chord of the circle. The perpendicular bisector of any chord is a line that contains a radius. Since the -axis is horizontal, any line perpendicular must be vertical. Using the midpoint formulas, the -coordinate of the midpoint of our chord must be  , hence the equation of the perpendicular bisector of the chord, and an equation of a line that contains the radius of the desired circle is then  .
Using similar analysis, there is a horizontal line,  , that also contains a radius.
Two different lines that each contain a radius of a given circle must, perforce, intersect at the center of the circle. The intersection of a vertical line whose points all have an -coordinate of 9 and a horizontal line whose points all have a  -coordinate of 7, must intersect in the point ) , which is therefore the center of the circle.
Since any of the given points are on the circle, the distance from the center to any of the points is the radius. In that you only need the radius squared to write your equation, you can use a modified form of the distance formula:
^2\ +\ (y_1\ -\ y_2)^2)
Once you know  all that is left is to plug in your numbers:
^2\ +\ \left(y\,-\,k\right)^2\ =\ r^2)
is a circle of radius  centered at ) .
John
My calculator said it, I believe it, that settles it
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