Intercept of a Line with a CircleDate: 05/28/99 at 06:24:11 From: Steven Doyle Subject: Intercept of a line with a circle Given the line, ax + by + c = 0 and a circle of arbitrary center and radius, x^2 + y^2 - 2fx - 2gy + d = 0 where, d = f^2 + g^2 - r^2 with, (f,g): Circle's Center r: Circle's Radius What is the general equation for the intercept between this circle and line? I know it should be some sort of quadratic answer (as you can have 0, 1, or 2 intercept points) but I can't seem to derive it. Date: 05/28/99 at 08:05:16 From: Doctor Jerry Subject: Re: Intercept of a line with a circle Hi Steven, A general equation can be found (see below; I did it with Mathematica), but I think that it is not very useful. In an attempt to make it a bit simpler (one doesn't have to worry about the possibility that a = 0 or b = 0), I used the parametric form of a line through points (a1,a2) and (b1,b2). The form is x = a1+t(b1-a1) y = a2+t(b2-a2). Replace x and y in the circle equation (x-h)^2+(y-k)^2-r^2 = 0 by the expressions x = a1+t(b1-a1) and y = a2+t(b2-a2) and then solve for t. In general, you'll obtain two values of t. Substitute these into the parametric equations for the intersection points. You can look just at the discriminant of the quadratic in the solution. If the discriminant is positive, there are two intersection points; if the discriminant is zero, just one (tangency); if the discriminant is negative, there is no intersection. t = (2*a1^2 + 2*a2^2 - 2*a1*b1 - 2*a2*b2 - 2*a1*h + 2*b1*h - 2*a2*k + 2*b2*k - Sqrt[(-2*a1^2 - 2*a2^2 + 2*a1*b1 + 2*a2*b2 + 2*a1*h - 2*b1*h + 2*a2*k - 2*b2*k)^2 - 4*(a1^2 + a2^2 - 2*a1*b1 + b1^2 - 2*a2*b2 + b2^2)* (a1^2 + a2^2 - 2*a1*h + h^2 - 2*a2*k + k^2 - r^2)])/ (2*(a1^2 + a2^2 - 2*a1*b1 + b1^2 - 2*a2*b2 + b2^2))} t = (2*a1^2 + 2*a2^2 - 2*a1*b1 - 2*a2*b2 - 2*a1*h + 2*b1*h - 2*a2*k + 2*b2*k + Sqrt[(-2*a1^2 - 2*a2^2 + 2*a1*b1 + 2*a2*b2 + 2*a1*h - 2*b1*h + 2*a2*k - 2*b2*k)^2 - 4*(a1^2 + a2^2 - 2*a1*b1 + b1^2 - 2*a2*b2 + b2^2)* (a1^2 + a2^2 - 2*a1*h + h^2 - 2*a2*k + k^2 - r^2)])/ (2*(a1^2 + a2^2 - 2*a1*b1 + b1^2 - 2*a2*b2 + b2^2)) - Doctor Jerry, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/