Conic Sections
    Circle and Line
      Condition of tangency - Condition for a line to be the tangent to a circle
         Condition for a line to be the tangent to the translated circle
      Tangents to a circle from a point outside the circle - use of the tangency condition examples
      Angle between a line and a circle
Condition of tangency - Condition for a line to be the tangent to a circle
Condition for a line to be the tangent to the circle with center at the origin O (0, 0)
A line touches a circle if the distance of the center of the circle to the line is equal to the radius of the circle, 
i.e., if
d = r.
The distance of the center S (0, 0) of a circle  x2 + y2 = r2  from a line  y = mx + c  or  - mx  + y - c = 0,
and after squaring obtained is   r2(m2 + 1) = c2   the condition for a line  y = mx + c  to be a tangent to 
the circle  x2 + y2  = r2.
Condition for a line to be the tangent to the translated circle  (x - p)2 + (y - q)2 = r2
In this case, the distance d of the center S(p, q) of the circle to a line  -mx  + y - c = 0 ( or  y = mx + c ), must be equal to the radius r of the circle, thus
after squaring obtained is   r2(m2 + 1) = (q -m p - c)2
the condition for a line  y = mx + c  to be a tangent to a translated circle with the center at S(p, q).
We can derive the same conditions using the tangency criteria which implies that the discriminant of the system of equations of a line and a circle, in that case, must be zero.
Then will the system have only one solution, i.e., the line and the circle will have only one common point, the tangency point. 
1)  The condition for a line  y = mx + c  to be tangent to the circle  x2 + y2  = r2,
    (1)    y = mx + c           plugging (1) into (2) gives the quadratic equation,
    (2)    x2 + y2 = r2                         (m2 + 1) x2 + 2mc x + c2 - r2 = 0   and according to the condition,
                               
D = b2 - 4ac = 0 or   (mc)2 - 4(m2 + 1)(c2 - r2) = 0 gives  r2(m2 + 1) = c2  the tangency condition.
2)  The condition for a line  y = mx + c  to be tangent to the circle  (x - p)2 + (y - q)2 = r2,
    (1)    y = mx + c
    (2)    (x - p)2 + (y - q)2 = r2
                                                      
by plugging (1) into (2) obtained is   (m2 + 1) x2 + 2(mc - mq - p) xp2 + c2 + q2 - r2 - 2cq = 0
then, from the condition  D = b2 - 4ac = 0 follows  r2(m2 + 1) = (q -m p - c)2  the tangency condition.
Tangents to a circle from a point outside the circle - use of the tangency condition
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 =>  (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
Conic sections contents
Copyright 2004 - 2020, Nabla Ltd.  All rights reserved.