Conic Sections
    Circle and Line
      Tangents to a circle from a point outside the circle - use of the tangency condition
      Angle between a line and a circle
Tangents to a circle from a point outside the circle - use of the tangency condition
Example:  Find the angle between tangents drawn from the point A(-1, 7) to the circle x2 + y2 = 25 .
Solution:   Equations of tangents we find from the system formed by equation of the line and the tangency condition,
  A(-1, 7) =>  (1)   y = mx + c  =>   c = m + 7  =>  (2)
                       (2)   r2(m2 + 1) = c2   
                                                      
            25(m2 + 1) = (m + 7)2
12m2 - 7m - 12 = 0 =>  m1 = - 3/4  and   m2 = 4/3
c1 = - 3/4 + 7 = 25/4  and   c2 = 4/3 + 7 = 25/3
therefore, the equations of tangents,
t1 ::  y- (3/4)x + 25/4  and  t2 ::  y =  (4/3)x + 25/3.
Slopes of tangents satisfy perpendicularity condition, that is
m1 = - 1/m2  =>    j = 90.
Example:   Find the area of the triangle made by points of contact of tangents, drawn from the point
A(15, 12) to the circle (x - 5)2 + (y - 2)2 = 20 , and the center S of the circle.
Solution:   From the system of equations formed by equation of the line through point A and the condition of 
tangency for the circle with the center at
S(p, q), we calculate slopes and intersections of tangents, thus 
A(15, 12) =>  (1)  y = mx + c,   c = 12 - 15m  =>  (2)
                     (2)   r2(m2 + 1) = (q -m p - c)2   
                                                                      
   S(5, 2)  and  r2  = 20   =>   (2)
20(m2 + 1) = (2 - 5m - 12 + 15m)2
2m2 - 5m + 2 = 0  =>   m1 = 1/2  and   m2 = 2,
 as  c = 12 - 15m  then  c1 = 9/2  and  c2 - 18
therefore, the equations of tangents,
t1 ::  y =  (1/2)x + 9/2   and   t2 ::  y =  2x - 18.
Coordinates of the tangency points we calculate by solving system of equations formed by equations of tangents and equation of the circle, thus
and the tangency point D2,
The area of the triangle SD1D2,
Example:   Given is a line  -3x + y + 1 = 0 and a circle x2 + y2 - 6x - 4y + 3 = 0, find equations of tangents to the circle which are perpendicular to the line.
Solution:   Slopes of tangents are determined by condition of perpendicularity, therefore
y = 3x - 1,    m  = 3  so that  mt  = -1/3
x2 + y2 -6x -4 y + 3 = 0  =>   (x - 3)2 + (y - 2)2 = 10
thus,   S(3, 2)  and  r2  = 10.
To find intersections c we use the tangency condition,
r2(m2 + 1) = (q -m p - c)2
10[(-1/3)2 + 1] = [2 -(-1/3)3 - c]or (3 - c)2 = 100/9
(3 - c) = 10/3  so  c1 = -1/3  and  c2 = 19/3.
The equations of tangents are,
t1 ::  y-(1/3)x -1/3  and  t2 ::  y-(1/3)x +19/3.
Example: Find equations of the common tangents to circles x2 + y2 = 13 and (x + 2)2 + (y + 10)2 = 117.
Solution:   Slopes and intersections of common tangents to the circles must satisfy tangency condition of both circles. Therefore, values for slopes m and intersections c we calculate from the system of equations,
The equation  -10 + 2m - c = +3c  does not satisfy given conditions.
The equation  -10 + 2m - c = -3c  or  c = 5 - m  plugged into (1)
Therefore, the equations of the common tangents are,
Angle between a line and a circle
The angle between a line and a circle is the angle formed by the line and the tangent to the circle at the intersection point of the circle and the given line.
Example:   Find the angle between a line 2x + 3y - 1 = 0 and a circle  x2 + y2 + 4x + 2y - 15 = 0.
Solution:   Coordinates of intersections of the line and the circle calculate by solving the system,
Rewrite the equation of the circle to standard form,
x2 + y2  + 4x + 2y - 15 = 0  Þ   (x - p)2 + (y - q)2 = r2
thus, (x + 2)2 + (y + 1)2 = 20S(-2, -1) and  r2  = 20.
Equation of the tangent at the intersection S1,
S1(-4, 3 =>    (x1 - p) (x - p) + (y1 - q) (y - q) = r2
(-4 + 2) (x + 2) + (3 + 1) (y + 1) = 20   =>    t ::   - x + 2y - 10 = 0  or   y = 1/2x + 5
The angle between the line and the circle is the angle formed by the line and the tangent to the circle at the intersection point, therefore
Intermediate algebra contents
Copyright 2004 - 2020, Nabla Ltd.  All rights reserved.