Conic Sections
    Circle and Line
      Line circle intersection
      Equation of a tangent at a point of a circle with the center at the origin
      Equation of a tangent at a point of a translated circle
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.
Equation of a tangent at a point of a translated circle (x - p)2 + (y - q)2 = r2
The direction vector of the tangent at the point P1(x1, y1), of a circle whose center is at the point S(p, q), and the direction vector of the normal, are perpendicular, so their scalar product is zero.
Points, O, S, P1 and P in the right figure, determine vectors,
Since their scalar product is zero, that is
 
Therefore,   is vector equation of a tangent at the point of a translated circle, 
or, when this scalar product is written in the component form,  (x1 - p) (x - p) + (y1 - q) (y - q) = r2
it represents the equation of the tangent at the point P1 (x1, y1), of a circle whose center is at S(p, q).
Example:  Find the angle formed by tangents drawn at points of intersection of a line x - y + 2 = 0 and
the circle 
x2 + y2 = 10.
Solution:  Solution of the system of equations gives coordinates of the intersection points,
Plug coordinates of  A and B into equation of the tangent:
Example:  At intersections of a line x - 5y + 6 =  0 and the circle x2 + y2 - 4x + 2 - 8 = 0 drown are tangents, find the area of the triangle formed by the line and the tangents.
Solution:  Intersections of the line and the circle are also tangency points. Solutions of the system of equations are coordinates of the tangency points,
(1)  x - 5y + 6 = 0     =>    x =  5 - 6    =>   (2)
(2)  x2 + y2 - 4x + 2 - 8 = 0
                                                
(5 - 6)2 + y2 - 4(5 - 6) + 2 - 8 = 0
 y2 - 3y + 2 = 0,    =>   y1 = 1  and   y2 = 2
                       x1 =  5 1 - 6 = -1,    =>   A(-1, 1),
                       x2 =  5 2 - 6 = 4,      =>    B(4, 2).
Rewrite the equation of the circle into standard form,
(x - p)2 + (y - q)2 = r2
   x2 + y2 - 4x + 2 - 8 = 0    =>   (x - 2)2 + (y + 1)2 = 13,   thus  S(2, -1) and  r = 13.
Plug tangency points A and B into equation of the tangent,
A( -1, 1)  and  B(4, 2)    =>   (x1 - p) (x - p) + (y1 - q) (y - q) = r2
(-1 - 2) (x - 2) + (1 + 1) (y + 1) = 13    =>    t1 ::   - 3x + 2y - 5 = 0,
 (4 - 2) (x - 2) + (2 + 1) (y + 1) = 13     =>    t2 ::    2x + 3y - 14 = 0.
Solution of the system of equations of tangents determines the third vertex C of the triangle,
  - 3x + 2y - 5 = 0    Tangents are perpendicular since their slopes satisfy the condition,  m1 = - 1/m2.     
   2x + 3y - 14 = 0,       C(1, 4)
                                 
The triangle ABC is right isosceles, whose area  A = 1/2 AC 2 = 1/2 (22 + 32)2 = 13/2 square units.
Conic sections contents
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