Equation of a circle on a plane

A circle

A circle is a geometrical locus of points, equidistant from one point - the center of the circle.
Let a point M with coordinates (a; b) be the center of a circle of radius R, and let a point N(x; y) be any point on this circle. Then, by the definition of radius, R=MN. Let the point N₁(x₁; y₁) lie inside the circle, the point N₂(x₂; y₂) lie outside the circle, and the point N(x; y) lies on the circle.

Equation of a circle

Using the formula for the length of a segment, we get By squaring the last equation, we obtain the equation of a circle: Only those points that belong to this circle satisfy this equation.
For example, the equation (x + 1)² + (y - 3)² = 25 defines a circle with center (-1;3) and radius R=√25=5.
If the center of the circle is at the origin, i.e. a=0, b=0, then the equation of the circle takes the form:

x² + y² = R².

For example, the equation x² + y² = 8 defines a circle with its center at the origin (0;0) and radius R=√8=2√2.

Conditions for the location of a point relative to a circle

Let us consider a circle defined by the equation (x - a)² + (y - b)² = R².
If the point N₁(x₁; y₁) lies inside the circle, then the distance from the center (a; b) of the circle to this point is less than the radius of the circle. Therefore, if the condition

(x₁ - a)² + (y₁ - b)² < R²,

is true, then the point N₁ is located inside the circle.
If the point N₂(x₂; y₂) lies outside the circle, then the distance from the center of the circle to this point is greater than the radius of the circle. Therefore, if the condition

(x₂ - a)² + (y₂ - b)² > R²,

is true, then the point N₂ is located outside the circle.
If the condition

(x - a)² + (y - b)² = R²,

is true for the point N(x;y), then the point N(x;y) lies on the circle.

Features of the circle equation

If we expand the brackets in the equation of a circle, we obtain the following second-order equation:

x² + y² - 2ax - 2by + a² + b² - R² = 0.

Thus, it is easy to see that the coefficients of x² and y² are equal to 1 and there is no term containing the product xy.
Consider a second-order equation of the form:

Ax² + Ay² + Bx + Dy + E = 0.

Let's divide both of its parts by A and select the perfect squares:

x² + y² + B/A⋅x + D/A⋅y + E/A = 0 ⇔

⇔ (x² + 2B/(2A)⋅x + B²/(4A²)) + (y² + 2D/(2A)⋅y + D²/(4A²)) + E/A - B²/(4A²) - D²/(4A²) = 0 ⇔

⇔ (x + B/(2A))² + (y + D/(2A))² = (B² + D² - 4AE) / (4A²).

Denoting B/(2A)=-a, D/(2A)=-b, the right side of the last equation (B²+D²-4AE) / (4A²) = F, we obtain the equation:

(x - a)² + (y - b)² = F.

Let us consider three cases for the value of the right side of the equation F: F>0, F=0 and F<0.
1) If F>0, then the equation defines a circle with center at point (a; b) and radius R=√F.
2) If F=0, then the equation takes the form (x - a)² + (y - b)² = 0 and defines a single point with coordinates (a; b).
3) If F<0, then the equation has no real roots. Its solutions will only be complex numbers, and the equation in this case is called the equation of an imaginary circle.

Examples of determining the coordinates of the center and radius of a circle given by a second-order equation

1) x² + y² - 10x + 4y + 20 = 0
This equation can be transformed as follows:

(x² - 10x + 25) + (y² + 4y + 4) + 20 - 25 - 4 = 0 ⇔ (x - 5)² + (y + 2)² = 9

The last equation defines a circle whose center is at the point (5;-2), and whose radius is R=3.

2) 3x² + 3y² + 6x - 12y + 8 = 0
This equation can be transformed as follows:

3(x² + 2x + 1) + 3(y² - 4y + 4) + 8 - 3 - 12 = 0 ⇔ 3(x + 1)² + 3(y - 2)² = 7

Dividing the last equation by 3, we get

(x + 1)² + (y - 2)² = 7/3

The last equation defines a circle whose center is at the point (-1;2), and whose radius is R=√7/3.

3) x² + y² - 16x - 2y - 15 = 0
Let's transform this equation into the form:

(x - 8)² + (y - 1)² = -50

This equation has no real roots and defines an imaginary circle.