Graphs of Equations in Two Variables

The graph of an equation in two variables $x$ and $y$ is the set of all points $(x, y)$ in the coordinate plane that satisfy the equation. Sketching a graph means drawing enough representative points and connecting them with a smooth curve to reveal the equation's main features.

Quick Reference

Shape	Equation	Key Parameters
Circle (standard form)	$(x - a)^{2} + (y - b)^{2} = R^{2}$	Center $(a, b)$ , radius $R$
Circle (general form)	$x^{2} + y^{2} + A x + B y + E = 0$	Requires ${(\frac{A}{2})}^{2} + {(\frac{B}{2})}^{2} - E > 0$

Graphs of Equations

An equation such as $y = \frac{x^{2}}{2} + 1$ connects two variables: $x$ and $y$ . An ordered pair satisfies the equation if it makes the equation true when substituted. For instance, $(0,1)$, $(- 1, 1.5)$ , $(1, 1.5)$, and $(2.1, 3.205)$ all satisfy $y = \frac{x^{2}}{2} + 1$ . There are infinitely many such pairs.

Some representative graphs are shown below.

An upward-opening parabola representing y = x²/2 + 1, with vertex at (0, 1). — (a) Graph of $y = \frac{1}{2} x^{2} + 1$

A straight line with positive slope representing y = 3x/4 − 1. — (b) Graph of $y = \frac{3}{4} x - 1$

A rightward-opening parabola representing y² = x. — (c) Graph of $y^{2} = x$

Circles

A circle with radius $R$ and center $C (a, b)$ is defined as the set of all points $P (x, y)$ in the plane at distance $R \geq 0$ from $C (a, b)$ . Applying the Distance Formula:

| P C | = \sqrt{(x - a)^{2} + (y - b)^{2}} = R

Squaring both sides gives:

Standard Form of the Equation of a Circle

(x - a)^{2} + (y - b)^{2} = R^{2}

where $R$ is the radius and $(a, b)$ is the center of the circle.

A circle with center C(a, b) and radius R. A point P(x, y) lies on the circle, connected to the center by a radius segment. — All points P(x, y) at distance R from C(a, b) satisfy $(x - a)^{2} + (y - b)^{2} = R^{2}$

Example 1. Write the equation of the circle with center $C (- 1, 1)$ that contains the point $P (- 4, 5)$ .

Solution. Since $P$ lies on the circle, the radius equals $| P C |$ :

R = | P C | = \sqrt{(- 4 - (- 1))^{2} + (5 - 1)^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5.

Substituting $a = - 1$ , $b = 1$ , $R = 5$ into the standard form:

(x + 1)^{2} + (y - 1)^{2} = 25.

Graph of the circle (x + 1)² + (y − 1)² = 25, centered at (−1, 1) with radius 5. — Graph of $(x + 1)^{2} + (y - 1)^{2} = 25$

General Form of a Circle

Starting from the standard form of a circle with radius 2 and center $(2, - 3)$ :

(x - 2)^{2} + (y + 3)^{2} = 4.

Expanding the left side and simplifying:

\begin{aligned} x^2 - 4x + 4 + y^2 + 6y + 9 &= 4 \\ x^2 + y^2 - 4x + 6y + 9 &= 0. \end{aligned}

This expanded form is called the general form of the circle equation.

General Form of the Equation of a Circle

x^{2} + y^{2} + A x + B y + E = 0

provided that ${(\frac{A}{2})}^{2} + {(\frac{B}{2})}^{2} - E > 0$ .

The condition ensures the right-hand side of the standard form is positive (i.e., a valid radius squared).

Finding Center and Radius by Completing the Square

To convert the general form to standard form, use the technique of completing the square. Here is the procedure:

Steps to Convert General Form to Standard Form

Group terms: Move the constant to the right and group the $x$ - and $y$ -terms: $x^{2} + y^{2} + A x + B y + E = 0 ⟹ (x^{2} + A x) + (y^{2} + B y) = - E$
Complete the square for each group by adding the square of half the linear coefficient to both sides:
- For $x$ : add ${(\frac{A}{2})}^{2}$ , so $x^{2} + A x = {(x + \frac{A}{2})}^{2} - {(\frac{A}{2})}^{2}$ .
- For $y$ : add ${(\frac{B}{2})}^{2}$ , so $y^{2} + B y = {(y + \frac{B}{2})}^{2} - {(\frac{B}{2})}^{2}$ .
Rewrite in standard form: ${(x + \frac{A}{2})}^{2} + {(y + \frac{B}{2})}^{2} = {(\frac{A}{2})}^{2} + {(\frac{B}{2})}^{2} - E$
Identify center and radius:
- Center: $(- \frac{A}{2}, - \frac{B}{2})$
- Radius: $R = \sqrt{{(\frac{A}{2})}^{2} + {(\frac{B}{2})}^{2} - E}$

Worked Examples

Example 2. Find the center and radius of the circle:

x^{2} + y^{2} - 12 x + 14 y + 69 = 0.

Solution. Group and move the constant to the right:

(x^{2} - 12 x) + (y^{2} + 14 y) = - 69.

Complete the square for each variable. For $x^{2} - 12 x$ , add ${(\frac{- 12}{2})}^{2} = 36$ . For $y^{2} + 14 y$ , add ${(\frac{14}{2})}^{2} = 49$ . Add 36 and 49 to both sides:

\begin{aligned} (x^2 - 12x + 36) + (y^2 + 14y + 49) &= -69 + 36 + 49 \\ (x-6)^2 + (y+7)^2 &= 16 \\ (x-6)^2 + (y-(-7))^2 &= 4^2. \end{aligned}

The center is $(6, - 7)$ and the radius is $4$.

Example 3. Show that $x^{2} + y^{2} + 4 x - 6 y + 9 = 0$ represents a circle, and find its center and radius.

Solution. Group and move the constant to the right:

(x^{2} + 4 x) + (y^{2} - 6 y) = - 9.

Complete the square: for $x^{2} + 4 x$ add ${(\frac{4}{2})}^{2} = 4$ ; for $y^{2} - 6 y$ add ${(\frac{- 6}{2})}^{2} = 9$ . Add 4 and 9 to both sides:

\begin{aligned} (x^2 + 4x + 4) + (y^2 - 6y + 9) &= -9 + 4 + 9 \\ (x+2)^2 + (y-3)^2 &= 4. \end{aligned}

Since the right-hand side is positive, this is indeed a circle. The center is $(- 2, 3)$ and the radius is $R = 2$ .

Frequently Asked Questions

What is the graph of an equation?

The graph of an equation in two variables is the set of all points $(x, y)$ whose coordinates satisfy the equation. For example, the graph of $y = x^{2}$ is a parabola because every point on the parabola, such as $(2, 4)$ or $(- 3, 9)$ , satisfies $y = x^{2}$ , and no other points do.

What is the standard form of a circle equation?

The standard form of a circle with center

(a, b)

and radius

R

is:

(x - a)^{2} + (y - b)^{2} = R^{2}

This form is convenient because you can read off the center and radius directly. For example,

(x - 3)^{2} + (y + 1)^{2} = 25

has center

(3, - 1)

and radius $5$.

What is the difference between the standard form and the general form of a circle?

The standard form

(x - a)^{2} + (y - b)^{2} = R^{2}

makes the center and radius immediately visible. The general form

x^{2} + y^{2} + A x + B y + E = 0

is the expanded version, where the center and radius are not obvious. To find the center and radius from the general form, complete the square on the

x

- and

y

-terms to convert it back to standard form.

How do I complete the square for a circle equation?

To complete the square for an expression like

x^{2} + A x

, take half the coefficient of

x

, square it, and add it to both sides of the equation:

x^{2} + A x + {(\frac{A}{2})}^{2} = {(x + \frac{A}{2})}^{2}

Repeat for the

y

-terms. The result is the standard form of the circle, from which you can identify the center and radius directly.

What does the condition

(A / 2)^{2} + (B / 2)^{2} - E > 0

mean?

In the general form

x^{2} + y^{2} + A x + B y + E = 0

, after completing the square the right-hand side of the standard form equals

{(\frac{A}{2})}^{2} + {(\frac{B}{2})}^{2} - E

. This must be strictly positive for the equation to represent a real circle with positive radius. If this expression equals zero, the "circle" degenerates to a single point. If it is negative, there are no real points satisfying the equation.

Graphs of Equations in Two Variables

Graphs of Equations in Two Variables

Quick Reference

Graphs of Equations

Circles

General Form of a Circle

Finding Center and Radius by Completing the Square

Worked Examples

Frequently Asked Questions

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