#### Answer

See below:

#### Work Step by Step

Consider the provided equation:
${{x}^{2}}+{{y}^{2}}=1$
The provided equation can also be written as:
${{\left( x-0 \right)}^{2}}+{{\left( y-0 \right)}^{2}}={{\left( 1 \right)}^{2}}$
Now, this represents the standard equation of a circle with its center at $\left( 0,0 \right)$ and radius of $1$ unit.
The graph for the equation ${{x}^{2}}+{{y}^{2}}=1$ can be plotted.
Now, consider the provided point:
$\left( -\frac{1}{2},\frac{\sqrt{3}}{2} \right)$
Check whether this point satisfies the provided equation by substituting the point in the provided equation:
Put the value of the x-coordinate and y-coordinate in the equation ${{x}^{2}}+{{y}^{2}}=1$
$\begin{align}
& {{\left( \frac{-1}{2} \right)}^{2}}+{{\left( \frac{\sqrt{3}}{2} \right)}^{2}}\overset{?}{\mathop{=}}\,1 \\
& \frac{1}{4}+\frac{3}{4}\overset{?}{\mathop{=}}\,1 \\
& 1=1
\end{align}$
The result is true. So, the provided point also lies on the curve.
The equation ${{x}^{2}}+{{y}^{2}}=1$ represents the standard equation of a circle with its center at $\left( 0,0 \right)$ and radius $1$ unit. The point $\left( -\frac{1}{2},\frac{\sqrt{3}}{2} \right)$ lies on the curve.