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.

### 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:

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

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

\begin{equation}

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

\end{equation}

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

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

\[x^{2}+y^{2}+Ax+Bx+E=0.\]