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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:
 Print circle equation summary
Solved examles:
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circle equations summary

Print circle equation 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)
(distance from the origin and angle)
 General form of a circle equation in polar form is obtained by using the law of cosines on the triangle that extends 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 Print example 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 distance is given by:
r=√((x_1-x_2 )^2+(y_1-y_2 )^2 )=√((3-2)^2+(4-1)^2 )=√10
And the equation of the circle is. (x-2)^2+(y-1)^2=10
Or after multiplying of the parenthesis. x^2+y^2-4x-2y-5=0
Example 1a - Points on circle Print example circle equation ▲
Find the points on the circle   (x − 2)2 + (y + 5)2 = 36   whose x is equal to 4.
First we will solve the general case of the circle:   (x + h)2 + (y + k)2 = R2  if we set the x point as xp then we have.
  (xp + h)2 + (y + k)2 = R2
After some simple algebric steps we get: (xp + h)2 + y2 + 2ky + k2 − R2 = 0
We get a quadratic equation of   y y2 + 2ky + (xp + h)2 + k2 − R2 = 0
And the solutions are: y_1,2=-k±√((x_p+h)^2+h^2-R^2 )
In our case   Xp = 4,   h = − 2   and  k = 5 y_1,2=-5±√32=0.65,-10.65
Example 2 - Circle normal line Print example circle normal line equation ▲
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 square  (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: m=dy/dx=(-3+2)/(2-0)=-0.5
And the line equation will be: m=(y-y_1)/(x-x_1 )=(y+3)/(x-2)
y + 3 = −0.5 ( x - 2)
x + 2y + 4 = 0
Example 3 - Circle tangent line when point is on the circle Print example circle tangent line ▲
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: 2x+2y dy/dx+3-4 dy/dx=0
dy/dx=m=(-2x-3)/(2y-4)=(-2∙(-4)-3)/(2∙1-4)=-2.5
And the tangent line equation is m=(y-y_1)/(x-x_1 )=(y-1)/(x+4)
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 perpendicular 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 m_c=(y_2-y_1)/(x_2-x_1 )=(2-1)/(-1.5+4)=1/2.5
Then the slope of the tangent line is: m=-1/m_c =-2.5
We get the same slope as in the first method.
Example 3a - Circle tangent line when point is on the circle Print example circle tangent line ▲
Find the equation of the tangent line to the circle  (x + 1)2 + (y − 2)2 = 5  and a point (1 , 3) which lies on the circle.

If we denote the center of the circle by the point   (h = −1 , k = 2)   then the circle eaution is:

(x − h)2 + (y − k)2 = R2

By implicit differentiation of the circle equation, we will find the slope of any tangent line to the circle.

The implicit derivation is: 2(x-h)+2(y-k) dy/dx=0
(x , y is the tangency point) dy/dx=m=(h-x)/(y-k)
And the tangent line equation is: (x1 , y1) is a point on the circle. y = mx +(y1 − mx1)
The slope of the tangent line is: dy/dx=m=(h-x)/(y-k)=(-1-1)/(3-2)=(-2)/1=-2
According to the tangent line equation we have   x1 = 1 and y1 = 3.
y = −2x + (3 + 2 · 1)
2x + y − 5 = 0
Example 4 - Circle shifted tangent line Print example circle 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 brought closer to the circle by kipping the same incline until it touches the circle.
Figure of example 4
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:
m_1=-1/m=-4
The angle θ is:       θ = tan-1(−4) = −75.96
The simplest 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:
mx-y-mx_t+y_t=0
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 tangent to x axis Print example circle tangent to x axis ▲
A circle's center is located at the point (2, -3). Find the equation of the circle that is tangent to the x axis, and also find the intersection points of the intersection of the circle with the  y  axis.
Figure of example 5
The center of the circle is at  (2 , −3)
The tangent point of the circle and the  x  axis is at  (2 , 0)
Hence the radius of the circle  r  is:
r=√((x_1-x_0 )^2+(y_1-y_0 )^2 )
r=√((2-2)^2+(-3-0)^2 )=3

Once we know the radius and the center of the circle, we have the equation of the circle as:

(x-x_c )^2+(y-y_c )^2=r^2
(x-2)^2+(y-(-3))^2=r^2
(x-2)^2+(y+3)^2=r^2

After multipling the paranthesis we get the equation of the line as:

x2 + y2 − 4x + 6y + 4 = 0

In order to find the intersection points with the y axis substitute  x = 0  in the circle's equation.

We get the quadratic equation:         y2 + 6y + 4 = 0

y=(-6±√(36-16))/2=(-6±√20)/2=-3±√5=-0.76,-5.24

And the intersection points are:     (0 , −0.76) and    (0 , −5.24)
Example 6 - Circle tangent to line Print example circle tangent to x axis ▲
Find the equation of a circle with center at  (2 , 3)  that is tangent to the line   3x + 4y + 2 = 0.
Figure of example 6
The slope of the line by explicit derivation is:  
3+4 dy/dx=0
dy/dx=m=-3/4
And the perpendicular line's slope  M (green line) is:
M=-1/m=4/3

The perpendicular line equation is:     y − yc = M (x − xc)

y-3=4/3 (x-2)

4x − 3y + 1 = 0

Solving this line equation and the given line equation to get the intersection point which is also located on the circle.

We get the solutions     x = −0.4    and    y = −0.2
The value of the radius is: r=√((2+0.4)^2+(3+0.2)^2 )=4
And the equation of the circle is: (x − 2)2 + (y − 3)2 = 16
After multipying paranthasis we get: x2 + y2 − 4x − 6y − 3 = 0
Example 7 - Circle tangent to line Print example circle tangent to x axis ▲
Find the equation of a circle which is tangent to both axii  x  and  y  at the 4th quadrant  (x > 0 , y < 0)  and has a radius of  6.
Figure of example 7
From the drawing we can see immidiatlly that the intersection points with the x and y axes are at points   (5 , 0) and (0 , −5).

And the radiuse is equal to 5.

Hence the equation of the circle is:

(x − 5)2 + (y + 5)2 = 52

After multiplying we get: x2 + y2 − 10x + 10y + 25 = 0
Example 8 - Circle tangent to 2 lines Print example circle tangent to x axis ▲
Find all the equations of the circles which are tangent to the  x  axis and the line   y = 2x   and have a radius of 6.
Figure of example 8
We notice from the drawing that there are four such circles that fulfills the conditions of the question.

If we denote the equation of a line as:

y = ax + b

The slope of the line is:

dy/dx=m=tan⁡α=a

The line angle is: α=tan^(-1)⁡a
The value of x1 is: x_1=2r/tan^(-1)⁡a
Circle center is at point: (2r/tan^(-1)⁡a ,r)

The equations of the upper right and lower left circles (blue circles) are:

(x-2r/tan^(-1)⁡a )^2+(y-r)^2=r^2 (1)   (2)

The equations of the lower left and upper right circles (green circles) are:
Notice that the angle now is  (π − α)/2)

(x+2r/tan^(-1)⁡a )^2+(y-r)^2=r^2 (3)   (4)

Notice that each circle center is located at a different quadrant.
Example 15 - Circle polar form Print example circle polar form ▲
Circle center is given by the polar coordinate to be (5 , pi/3). Find the equation of the 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.
Figure of example 5
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 + r02 − 2dr0cos(θ − φ)
In the example we get:     r0 = 5,      φ = π/3    and    R = 2.
Substituting these values to the circle polar form we get:
d^2+25-2∙5d cos⁡(θ-π/3)=4
d^2-10d cos⁡(θ-π/3)+21=0
Figure Case Circle polar equation
Example solution
Circle center at x axis circle center on the  x  axis  (φ = 0)

example:     (x − r0)2 + y2 = R2		 d2 − 2dr0 cosθ + r02 − R2 = 0
d2 − 10d cosθ + 21 = 0
Circle center at x axis circle center on the  x  axis  (φ = 0) (r0 = R)

example:     (x − r0)2 + y2 = R2		 d − 2R cosθ = 0
d − 4 cosθ = 0
Circle center at y axis circle center on the  y  axis  (φ = 90°)

example:     x2 + (y − r0)2 = R2		 d2 − 2dr0 sinθ + r02 − R2 = 0
d2 − 10d sinθ + 21 = 0
Circle center at y axis circle center on the  y  axis  (φ = 90°) (r0 = R)

example:     x2 + (y − r0)2 = R2		 d − 2Rd sinθ = 0
d − 4 sinθ = 0
Circle passing through the origin circle passes through the origin  (r0 = R) d − 2R cos(θ − φ) = 0
d=10 cos⁡(θ-π/3)