# Exponential graphs and circles

## â€‹â€‹In a nutshell

Equations of lines and curves show relationships. One such relationship is an exponential relationship, which tracks a very small positive number to infinity, or vice versa. Some curves look like shapes, for example, circles can be drawn using equations. These involve quadratic terms in *both* $x$ and $y$.

## Exponential graphs

Exponential equations have the form $A^x$ or $A^{-x}$ where $A$ is a positive constant.

**Note: ***â€‹$A^{-x}=\frac1{A^x}$ which is the same as $(\frac1A)^x$. So really, $A^x$ and $A^{-x}$ do not actually represent different types of exponential graphs.*

An example of an exponential graph is given below. This is $y=2^x$:â€‹

If $A$ is bigger than $1$, then the exponential graph will have a very similar shape. They all pass the $y$-axis at $1$, and never touch the $x$-axis. Another example is given below. This is $y=(\frac12)^x$ (which is the same as $y=2^{-x}$):â€‹

If $A$ is between $0$ and $1$, it will have this shape. Notice that the curve still passes through $(0,1)$ and never touches the $x$-axis. The difference is that it is reflected in the $y$-axis.

Exponential graphs span the whole $x$-axis, so any $x$-coordinate inserted into an exponential equation will return a $y$-coordinate. This is not true of $y$-coordinates: notice that there are no points on the graph such that $y\leq0$.

## Circles

Circles can be graphed using the general equation

â€‹$x^2+y^2=r^2$

â€‹where $r$ is a constant and is the radius of the circle.

**Note:*** Circles using this equation will have their centre at the origin.*

An example is given below:

The equation of this circle is $x^2+y^2=1$ since the radius is $1$ and $1^2=1$.

## Finding the gradient of a tangent to a circle

A tangent is a line that touches a curve at one place, then continues on. For example, a tangent to a curve is given below:

You say that the tangent is at point $A$â€‹. Since tangents are straight lines they have the same gradient everywhere. At the point where a tangent touches the circle the tangent is perpendicular to the radius. To calculate the gradient of the tangent follow the procedure below:

#### Procedure

**1.**
| Identify the point on the circle where the tangent touches. Call this point $A$ with coordinates $(x_A,y_A)$â€‹. |

**2.**
| Calculate the gradient of the radius to point $A$â€‹. This will be $\frac{y_A}{x_A}$ (using $m=\frac{\text{change in }y}{\text{change in }x}$) since the centre is at the origin.â€‹ |

**3.**
| The gradient of the tangent is the negative reciprocal of this. This is $-\frac{1}{(\frac{y_A}{x_A})}=-\frac{x_A}{y_A}$â€‹. |

**Note:** *The reciprocal of a fraction is the same fraction but with the numerator and the denominator swapped.*

##### Example

*Find the gradient of the tangent at point $A(4,3)$â€‹ on the circle with equation*

*$x^2+y^2=25$*

*You already have the point where the tangent touches the circle. Now the gradient of the radius. This is $\frac34$. The gradient of the tangent is the negative reciprocal of this. This is $-\frac1{(\frac34)}=\underline{-\frac43}$â€‹.â€‹*