Randy Haas <firstname.lastname@example.org> wrote: >Does any one know of the equation of a circle that goes through 2 points >and is tangent to a vertical line. There will be 2 circles that solve the >problem. > >| >| * >| * >| >| > >Thanks for any help >Randy Haas
If we call the points P1 and P2, then we have two equations right off the start:
If we expand the squares and subtract one equation from the other, we get
(3) (x1-x2)h + (y1-y2)k = 0
The center of the circle (h,k) lies on the perpendicular bisector of the line through P1 and P2. If we let P3 = ((x1+x2)/2,(y1+y2)/2), the center of this line segment, the equation of the perp. bisector is
(y-y3)/(x-x3) = -(x2-x1)/(y2-y1) = -1/m
where m is the slope of the line joining P1 and P2.
Plugging (h,k) and rearranging a bit, we have
(4) h + mk = x3 + my3
We may without loss of generality assume the circle is tangent to the y-axis. If not, just substitute h = h+v at the end, if the vertical line was x=v. So,
h = r
Plugging that into (4) and (3), we get a quadratic in r:
(5) 1/m^2 r^2 - 2(y1/m + x1)r + (x1^2 + y1^2) = 0
That will give us 2 values for r, as we expected. From that it is possible (note: I did not say easy) to compute h and k.
I'll let you do that, and maybe simplify things if I've let things get out of hand. I may have missed some simplifying assumption.
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