Conic Sections  Circle and Line
Line circle intersection
A line and a circle in a plane can have one of the three positions in relation to each other, depending on the distance d of the center S(p, q) of the circle (x - p)2 + (y - q)2 = rfrom the line Ax + By + C = 0, where the formula for the distance: If the distance of the center of a circle from a line is such that: d < r,  then the line intersects the circle in two points, d = r,  the line touches the circle at only one point, d > r,  the line does not intersect the circle, and they have no common points. Example:  At which points the line  x + 5y + 16 = 0  intersects the circle  x2 + y2 - 4x + 2y - 8 = 0.
Solution:   To find coordinates of points at which the line intersects the circle solve the system of equations:  So, the line intersects the circle at points,  A(4, -4) and  B(-1, -3).
Example: Find equation of a circle with the center at S(1, 20) which touches the line 8x + 15y - 19 = 0.
Solution: If a line touches a circle then the distance between the tangency point and the center of the circle
d = DS  = r  i.e., thus, equation of the circle  (x - 1)2 + (y - 20)2 = 289. We can use another method to solve this problem. Since, the normal n through the center is perpendicular to the tangent t then the direction vector sn is perpendicular to the direction vector st . Therefore, as mt = sy/sx = -8/15 then  so, equation of the normal is As the tangency point D is the common point of the tangent and the normal then, putting coordinates of the
radius vector of the normal into equation of the tangent determines a value of the parameter
l to satisfy that
condition, as then, these variable coordinates of the radius vector put into equation of the tangent follows    8 · (1 + 8l) + 15 · (20 + 15l) - 19 = 0   =>     289l = 289   =>    l = - 1
 so, the radius vector of the tangency point therefore the tangency point  D(-7, 5). The radius of the circle, since This result we can check by plugging the coordinates of the tangency point into equation of the circle, that is
D(-7, 5)  =>   (x - 1)2 + (y - 20)2 = 289,    (-7 - 1)2 + (5 - 20)2 = 289   =>  (-8)2 + (- 15)2 = 289
therefore, the tangency point is the point of the circle.
Equation of a tangent at a point of a circle with the center at the origin
The direction vector of the tangent at the point P1 of a circle and the radius vector of P1 are perpendicular to each other so their scalar product is zero.
Points, O, P1 and P in the right figure, determine vectors,  the scalar product written in the components gives, x1x + y1y = r2
This is equation of a tangent at the point P1(x1, y1) of a circle with the center at the origin    College algebra contents E 