Example: The above circle is defined by the equation: (x - 7)^{2} + (y - 5)^{2} = r^{2}

Example: Change circle equation x^{2} + y^{2} - 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: (r_{0} , 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 r_{0}) 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:

Find the equation of the line normal to the circle x^{2}+ y^{2}+ 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}= a^{2}+ 2ab + b^{2}.

x^{2} + ( y^{2}+ 4y ) − 1 = 0

x^{2} + ( y + 2 )^{2} − 4 − 1 = 0

x^{2} + ( 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 x^{2}+ y^{2}+ 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 perpendicular to this slope, hence equal to minus the reciprocal of the slope.

The center of the circle is at (see example 2)

(x^{2}+ 3x) + (y^{2} −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 3a - Circle tangent line when point is on the circle

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.

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 simplest way to find the tangency point is by using trigonometry, notice that there are two tangency points on either side of the circle.

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.

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:

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

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

x^{2}+ y^{2} − 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: y^{2}+ 6y + 4 = 0

And the intersection points are: (0 , −0.76) and (0 , −5.24)

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.

In polar form the circle center is given by the value of the distance from the origin r_{0} to the circle center and the angle of this line (r_{0} , φ).

Applying the law of cosines on triangle R, d and r_{0} we have:

R^{2}= d^{2}+ r_{0}^{2} − 2dr_{0}cos(θ − φ)

In the example we get: r_{0} = 5, φ = π/3 and R = 2.

Substituting these values to the circle polar form we get: