Intersection of line and circle

[minase]

I'm trying to find the points of intersection
of line and circle with equations:

(x-p)^2 + (y-q)^2 = r^2
(y-y1)*(x2-x1)-(x-x1)*(y2-y1)=0

but i can't handle with this. Can anyone help me?

[ComputerGeek]

I'm trying to find the points of intersection
of line and circle with equations:
(x-p)^2 + (y-q)^2 = r^2
(y-y1)*(x2-x1)-(x-x1)*(y2-y1)=0
but i can't handle with this. Can anyone help me?

What do you know about the circle's dimensions?

[mathman]

Your second equation (the line) can be simplified to look like y=mx+b (I assume x1, x2,y1,y2 are constants). Substitute mx+b for y in the first equation. You now have a quadratic in x. Solve for x, 2 real roots gives points of intersection, 1 root is tangency point, 0 real roots means no intersection. If x roots are real, use second equation to get y values.

Special case x1=x2, then x comes right out of second equation and 2 values of y can be found. Above comments about real roots and intersections apply. Line happens to be vertical.

[SachinKainth]

Your second equation (the line) can be simplified to look like y=mx+b (I assume x1, x2,y1,y2 are constants). Substitute mx+b for y in the first equation. You now have a quadratic in x. Solve for x, 2 real roots gives points of intersection, 1 root is tangency point, 0 real roots means no intersection. If x roots are real, use second equation to get y values.

Special case x1=x2, then x comes right out of second equation and 2 values of y can be found. Above comments about real roots and intersections apply. Line happens to be vertical.

I see how the case where the line is not vertical works but I don't see what you would do in the case where it is vertical. Can you explain in a little more detail?