An equation such as $y=\frac{x^{2}}{2}+1$ connects two variables: $x$ and $y$. We say an ordered pair satisfy the equation if it makes the equation true when the values for $x$ and $y$ are substituted into the equation. For example, $(0,1),(-1,1.5),(1,1.5),(2.1,3.205),\ldots$ satisfy the given equation. There are infinite number of ordered pairs that satisfy such an equation.

The graph of an equation between $x$ and $y$ (or any other two variables) is the set of all points $(x,y)$ satisfying the equation. The phrase “plot the graph of an equation” or “sketch the graph of an equation” means to draw enough number of points and connect them by a smooth curve to illustrate the main features of the graph. Graphs of some equations are sketched in Figure 1.

 (a) Graph of $y=\frac{1}{2}x^{2}+1$ (b) Graph of $y=\frac{3}{4}x-1$ (c) Graph of $y^2=x$

Figure 1

### Circle

Now let’s find the equation of a circle with radius $r$ and center $C(a,b)$. By definition, this circle is the set of all points $P(x,y)$ in a plane that are at a distance $R\ge0$ from the point $C(a,b)$ (see the following figure). From the Distance formula, we have

$|PC|=\sqrt{(x-a)^{2}+(y-b)^{2}}=R.$

By squaring both sides, we obtain:

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

The standard form of the equation of a circle with radius $R$ and center $(a,b)$ is

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

 Figure 2: All points $P(x,y)$ that are at a distance $R$ from $C(a,b)$ satisfy the equation $(x-a)^{2}+(y-b)^{2}=R^{2}.$

Example

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

Solution

Because $P$ is on the circle, the radius $r$ is $|PC|$ (also denoted by $d(P,C)$). By the Distance formula, we have

$R=|PC|=\sqrt{(-4-(-1))^{2}+(5-1)^{2}}=\sqrt{25}=5.$

Substituting $a=-1,b=2$, and $r=5$ in Equation

gives us

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

This circle is illustrated in the following figure.

 Figure 3: Graph of $(x+1)^{2}+(y-1)^{2}=25$.

The standard form of the equation of a circle with radius 2 and center $(2,-3)$ is

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

We can change the look of this equation by expanding the terms on the left hand side (using the formula for the Square of a Sum on this page):

$x^{2}-4x+4+y^{2}+6y+9=4$

or

$x^{2}+y^{2}-4x+6x+13=4$

Subtracting 4 from both sides, the equation of the circle becomes

$x^{2}+y^{2}-4x+6x+9=0.$

The above equation is called the general form of the equation of the circle.

The general form of the equation of a circle is

$x^{2}+y^{2}+Ax+Bx+E=0.$