Equation of a Figure

In coordinate geometry, any geometric figure — a line, circle, parabola, or more complex curve — can be described by an algebraic equation. This connection between geometry and algebra is the foundation of analytic geometry.

What is the Equation of a Figure?

📐Definition — Equation of a Figure

An equation $F(x, y) = 0$ is called the equation of a figure $\Phi$ if the following two conditions both hold:

Completeness: Every point of $\Phi$ satisfies the equation, i.e., if $(x_0, y_0) \in \Phi$ then $F(x_0, y_0) = 0$ .
Sufficiency: Every point whose coordinates satisfy the equation belongs to $\Phi$ , i.e., if $F(x_0, y_0) = 0$ then $(x_0, y_0) \in \Phi$ .

In other words, the figure $\Phi$ is precisely the solution set of the equation $F(x, y) = 0$ in the coordinate plane.

Remark. A single equation $F(x,y) = 0$ typically defines a curve (a one-dimensional set of points). Two simultaneous equations generally define isolated points (intersections of two curves).

Locus of Points

A locus is the set of all points in the plane that satisfy a given geometric condition. The equation of a figure is the algebraic expression of the condition defining the locus.

Standard example: The locus of all points at distance $r$ from a fixed center $O$ is a circle of radius $r$ . We will now derive its equation.

Equation of a Circle

⚡Theorem — Circle Equation (Theorem 10.1)

The equation of the circle with center $(a, b)$ and radius $r > 0$ is:

$\boxed{(x - a)^2 + (y - b)^2 = r^2}$

Proof. A point $P(x, y)$ lies on this circle if and only if its distance from the center $(a, b)$ equals $r$ :

$|CP| = r \iff \sqrt{(x - a)^2 + (y - b)^2} = r \iff (x-a)^2 + (y-b)^2 = r^2 \qquad \blacktriangleleft$

Special cases:

Unit circle (center at origin, radius 1): $x^2 + y^2 = 1$
Circle centered at origin with radius $r$ : $x^2 + y^2 = r^2$

Expanding the Circle Equation

The standard form $(x-a)^2 + (y-b)^2 = r^2$ can be expanded:

$x^2 - 2ax + a^2 + y^2 - 2by + b^2 = r^2$

$x^2 + y^2 - 2ax - 2by + (a^2 + b^2 - r^2) = 0$

Setting $D = -2a$ , $E = -2b$ , $F = a^2 + b^2 - r^2$ , this takes the form:

$x^2 + y^2 + Dx + Ey + F = 0$

Conversely, any equation of this form (with $D^2 + E^2 - 4F > 0$ ) represents a circle. Completing the square recovers the center and radius.

Worked Examples

✎Example — Example 1 — Writing a Circle Equation

Write the equation of the circle with center $K(-2, 3)$ and radius $5$ .

Solution. Substitute $a = -2$ , $b = 3$ , $r = 5$ into the standard form:

$(x - (-2))^2 + (y - 3)^2 = 5^2$

$(x + 2)^2 + (y - 3)^2 = 25$

Answer: $(x + 2)^2 + (y - 3)^2 = 25$ . $\blacktriangleleft$

✎Example — Example 2 — Finding Center and Radius

Find the center and radius of the circle $x^2 + y^2 - 6x + 4y - 3 = 0$ .

Solution. Complete the square in $x$ and $y$ separately:

$\underbrace{(x^2 - 6x + 9)}_{(x-3)^2} + \underbrace{(y^2 + 4y + 4)}_{(y+2)^2} = 3 + 9 + 4 = 16$

$(x - 3)^2 + (y + 2)^2 = 16$

Comparing with $(x-a)^2 + (y-b)^2 = r^2$ : center $C(3, -2)$ and radius $r = 4$ .

Answer: Center $(3, -2)$ , radius $4$ . $\blacktriangleleft$

✎Example — Example 3 — Locus Problem

Find the locus of points equidistant from the points $A(0, 0)$ and $B(4, 0)$ .

Solution. Let $P(x, y)$ be any point with $|PA| = |PB|$ . Then:

$\sqrt{x^2 + y^2} = \sqrt{(x-4)^2 + y^2}$

Squaring both sides:

$x^2 + y^2 = (x-4)^2 + y^2 = x^2 - 8x + 16 + y^2$

$0 = -8x + 16 \implies x = 2$

The locus is the vertical line $x = 2$ , which is the perpendicular bisector of segment $AB$ . $\blacktriangleleft$

Exercises

✏Exercise

Write the equation of each circle.

Center $(0, 0)$ , radius $7$
Center $(3, -4)$ , radius $\sqrt{5}$
Center $(-1, 6)$ , passing through the origin $O(0,0)$

Hint for (3): The radius equals the distance from the center to the origin.

✏Exercise

Find the center and radius of each circle.

$x^2 + y^2 - 4x - 10y + 20 = 0$
$x^2 + y^2 + 8x - 6y = 0$
$(x + 1)^2 + y^2 = 9$

✏Exercise

A circle passes through the three points $A(1, 0)$ , $B(-1, 0)$ , and $C(0, 2)$ . Write its equation.

Hint: Substitute each point into $x^2 + y^2 + Dx + Ey + F = 0$ to obtain a system of three equations, then solve for $D$ , $E$ , $F$ .