Circle and Line

Condition for a line to be the tangent to the circle with the center at the origin O(0, 0)
Condition for a line to be the tangent to the translated 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 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:   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]2  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 = 20,  S(-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
Pre-calculus contents H