Conic Sections
    Circle and Line
      Line circle intersection
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.
Intermediate algebra contents
Copyright 2004 - 2020, Nabla Ltd.  All rights reserved.