Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
Drexel University or The Math Forum.



Re: Circle equation
Posted:
Jul 16, 1997 10:01 AM


Randy Haas <randychaas@msn.com> 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:
(1) (x1h)^2 + (y1k)^2 = r^2 (2) (x2h)^2 + (y2k)^2 = r^2
If we expand the squares and subtract one equation from the other, we get
(3) (x1x2)h + (y1y2)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
(yy3)/(xx3) = (x2x1)/(y2y1) = 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 yaxis. 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.
Steve Monson  Q: Why did Arnold Schwartzenegger and Maria Shriver get married? A: They're trying to breed a bulletproof Kennedy.



