Question

Show that the point is on the unit circle.

$\left(\frac{\sqrt{11}}{6},\frac{5}{6}\right)$

Short answer

The point$\left(\frac{\sqrt{11}}{6},\frac{5}{6}\right)$ is on the unit circle.

Step-by-step solution

Step 1. State the definition of the unit circle.

The unit circle is the circle of radius $1$ centered at the origin in the $xy$- plane. It is given by the equation $x^2+y^2=1$.

Step 2. Substitute the points.

In order to show that the point satisfies the equation of the unit circle, substitute $\frac{\sqrt{11}}{6}$ for $x$ and $\frac{5}{6}$ for $y$ into the equation $x^2+y^2=1$.

$\begin{array}{r}{x}^2+y^2\stackrel{?}{=}1\\ {\left(\frac{\sqrt{11}}{6}\right)}^2+{\left(\frac{5}{6}\right)}^2\stackrel{?}{=}1\end{array}$

Step 3. Simplify and state the conclusion.

Simplify the expression.

$$\begin{gathered} {\left(\frac{\sqrt{11}}{6}\right)}^2+{\left(\frac{5}{6}\right)}^2\stackrel{?}{=}1 \\ \frac{11}{36}+\frac{25}{36}\stackrel{?}{=}1 \\ \frac{36}{36}\stackrel{?}{=}1 \\ 1=1\text{True} \end{gathered}$$

Therefore, the point is on the unit circle.