## Precalculus (6th Edition) Blitzer

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.