The discussion revolves around finding the points of intersection between a line and a circle, represented by their respective equations. The circle is defined by the equation \((x-p)^2 + (y-q)^2 = r^2\), while the line is given in a general form that can be simplified to slope-intercept form.
The conversation is active, with participants exploring different methods to approach the problem. Some guidance has been provided regarding the substitution method and the interpretation of roots, but there is still a request for clarification on the vertical line case.
Participants are working under the assumption that the constants in the line's equation are known, but there is a lack of specific information about the circle's dimensions, which may affect the discussion.
minase said: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?
mathman said: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.