Conic Sections

General equation of a circle with the center S(p, q) - translated circle
The equation of the circle, example
Equation of the circle with the center at the origin O(0, 0)

The equation of the circle through three points, example

Circle
General equation of a circle with the center S(p, q) - translated circle
A circle with the center at the point S(p, q) and radius r is a set of all points P(x, y) of a plane to whom the
 distance from the center,   SP = r  or
is the general equation of a circle with the center S(p, q).
 Equation of the circle with the center at the origin O(0, 0),
Example: A circle passes through points A(2, 4) and B(-2, 6) and its center lies on a line x + 3y - 8 = 0.
Find equation of the circle.
Solution:   The intersection of the chord AB  bisector and the given line is the center S of the circle, since the bisector is normal through the midpoint M, then
 As the bisector is perpendicular to the line AB Equation of the bisector M(0, 5) and mn = 2  =>   y - y1 = m( x - x1), gives       y - 5 = 2(x - 0)   or   -2x + y - 5 = 0.
So, the equation of the circle,     (x - p)2 + (y - q)2 = r2   =>    (x + 1)2 + (y - 3)2 = 10.
Circle through three points
A circle is uniquely determined by three points not lying on the same line. If given are points, A, B and C then the intersection of any pair of the perpendicular bisectors of the sides of the triangle ABC is the center of the circle.
 Since all three points lie on the circle, their coordinates must satisfy equation of the circle (x - p)2 + (y - q)2 = r2. Thus, we obtain the system of three equations in three unknowns p, q and r. Subtracting second equation from first and then third from first we obtain two equations in two unknowns p and q. Solutions of that system plug into any of three equations to get r.
Example:  Find equation of a circle passing through three points, A(-2, -6), B(5, -7) and C(6, 0).
Solution:   The coordinates of the points, A, B and C plug into equation  (x - p)2 + (y - q)2 = r2,
Thus, the equation of the circle through points A, B and C,    (x - 2)2 + (y + 3)2 = 25.
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 =  5y  - 6  =>   (2) (2)  x2 + y2 - 4x + 2y  - 8 = 0 (5y  - 6)2 + y2 - 4(5y  - 6) + 2y  - 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 = Ö13.
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 · Ö(22 + 32)2 = 13/2 square units.
Pre-calculus contents H