﻿ Circle equation geometry
 Circle equation ▲
Circle equation ( x - )2 + ( y - )2 = 2
x2 + y2 + x + y + = 0
 r Circle center: ( , )
Circle area:
Circle circumference:
Point on circle:
 xp: y1 y2 yp: x1 x2
Circle polar form:
Polar circle center: ( , )
 Input limit:

 Example 1 Example 2 Example 3 Example 4 Example 5
 circle equations summary ▲
Circle equation (cartesian form) Example: The above circle is defined by the equation: (x - 7)2 + (y - 5)2 = r2

Example: Change circle equation
x2 + y2 - 6x + 4y -3 = 0 to form (1). Center of the circle is at (3, -2) and circle equation is:
(x - 3)2 + (y + 2)2 = 16 This is the equation of a circle with center at (a, b) and radius r. Relations between the coefficients of equations (1), (2), (3) and (4) are:
 (2) → (1) (1) → (2)  (4) → (3) (3) → (4) (4) → (1) (3) → (1) Circle equation (polar form) In polar coordinates a point is
described by: (r0 , φ)
(diatance from origin , angle)
General form of a circle equation in polar form is obtained by using the law of cosines on the triangle that extandes from the origin to the center of the circle (radius r0) and to a point on the circle (radius r) and back to the origin (side d). (5)
If the center of the circle lies on the x axes then the circle equation becomes: (6)
If the center of the circle lies on the y axes then the circle equation is: (7)
 Example 1 - Circle equation ▲
Find the equation of a circle with a center at (2 , 1) and passing through the point (3 , 4).
Because the given point lay on the circle then the distance between the circle center and the point is the radius. The distanance is given by: Or after multiplying of the parenthesis. Example 2 - Circle normal line ▲
Find the equation of the line normal to the circle  x2 + y2 + 4y −1 = 0  and passing through the point
(2 , −3).
First we have to find the equation of the line stretching from the center of the circle to the given point. The center of the circle can be found by the procedure for completing the squre  (a + b)2 = a2 + 2ab + b2.
 x2 + ( y2 + 4y ) − 1 = 0 x2 + ( y + 2 )2 − 4 − 1 = 0 x2 + ( y + 2 )2 = 5
This is a circle with a radius of square root of 5 and center at (0 , −2 ).
 The slope of the line is: And the line equation will be: y + 3 = −0.5 ( x - 2) x + 2y + 4 = 0
 Example 3 - Circle tangent line when point is on the circle ▲
Find the equation of the tangent line to the circle  x2 + y2 + 3x −4y −1 = 0  and a point (−4 , 1) which lies on the circle.
The solution of this problem can be solved in different ways we will show two methods. First by implicit differentiation of the circle equation, this will give us the slope of any point on the circle and then find the equation of the line by the slope and the given point.
 The implicit derivation is:  And the tangent line equation is m(x + 4) + 1 − y = 0 5x + 2y + 18 = 0
The second method is to find the slope of the line connecting the center of the circle to the given point and then the tangent line slope is perpendiculat to this slope hence equal to minus the reciprocal of the slope.
 The center of the circle is at (see example 2) (x2 + 3x) + (y2 −4y) − 1 = 0 (x + 1.5)2 + (y −2) = 7.25 This is the point (−1.5 , 2) the slope of the line from the center to the point Then the slope of the tangent line is: We get the same slope as in the first method.
 Example 4 - Circle shifted tangent line ▲
A circle is given by the equation  (x − 2)2 + (y + 1)2 = 9.  A line x −  4y + 15 = 0  is drawn outside the circle. Find the tangency point if the line is brougth closer to the circle by kipping the same incline until it touch the circle. The center of the circle is at  (2 , −1)
The slope of the given line is:    m = 1 / 4 = 0.25
The slope of the perpendicular radius is given by: The angle θ is:       θ = tan-1(4) =75.96
The most simple way to find the tangency point is by using trigonometry, notice that there are two tangency points on either side of the circle.
 xt = xc ± r cosθ = 2 ± 3 cos(−75.96) = 2.73 , 1.27 yt = yc ± r sinθ = −1 ± 3 sin(−75.96) = −3.91 , 1.91
And the two tangency points are:
(2.73 , −3.91)   and   (1.27 , 1.91).
The tangent line equations are: First point (2.73 , −3.91) Second point (1.27 , 1.91)
 0.25x − y − 0.25∙1.27 + 1.91 = 0 x − 4y − 1.27 + 7.64 = 0 And the 1st tangent line equation is: x − 4y + 6.37 = 0
 0.25x − y − 0.25∙2.73 − 3.91 = 0 x − 4y − 2.73 − 15.64 = 0 And the 2nd tangent line equation is: x − 4y − 18.37 = 0
 Example 5 - Circle polar form ▲
Circle center is given by the polar coordinate to be (5 , pi/3). Find the equation of thr circle if the radius is  2. Investigate the cases when circle center is on the  x  axis and second if it is on the  y  axis and in the origin. In polar form the circle center is given by the value of the distance from the origin  r0  to the circle center and the angle of this line  (r0 , φ).
Applying the law of cosines on triangle  R, d  and  r0  we have:
R2 = d2 + r022dr0cos(θ − φ)
In the example we get:     r0 = 5,      φ = π/3    and    R = 2.
Substituting this values to the circle polar form we get:  Figure Case Circle polar equationExample solution circle center on the  x  axis  (φ = 0) example:     (x − r0)2 + y2 = R2 d2 − 2dr0 cosθ + r02 − R2 = 0d2 − 10d cosθ + 21 = 0 circle center on the  x  axis  (φ = 0) (r0 = R) example:     (x − r0)2 + y2 = R2 d − 2R cosθ = 0d − 4 cosθ = 0 circle center on the  y  axis  (φ = 90°) example:     x2 + (y − r0)2 = R2 d2 − 2dr0 sinθ + r02 − R2 = 0d2 − 10d sinθ + 21 = 0 circle center on the  y  axis  (φ = 90°) (r0 = R) example:     x2 + (y − r0)2 = R2 d − 2Rd sinθ = 0d − 4 sinθ = 0 circle passes through the origin  (r0 = R) d − 2R cos(θ − φ) = 0 