Answer
The point $(0,1)$ is on the graph of the equation.
The point $\Big(\dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{2}}\Big)$ is on the graph of the equation.
The point $\Big(\dfrac{\sqrt{3}}{2},\dfrac{1}{2}\Big)$ is on the graph of the equation.
Work Step by Step
$x^{2}+y^{2}=1;$ $(0,1)$ $,$ $\Big(\dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{2}}\Big)$ $,$ $\Big(\dfrac{\sqrt{3}}{2},\dfrac{1}{2}\Big)$
Substitute each point into the equation:
$(0,1)$
$x^{2}+y^{2}=1$
$0^{2}+1^{2}=1$
$1=1$ True
Substituting the point $(0,1)$ into the equation makes it true. This point is on the graph of the equation.
$\Big(\dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{2}}\Big)$
$x^{2}+y^{2}=1$
$\Big(\dfrac{1}{\sqrt{2}}\Big)^{2}+\Big(\dfrac{1}{\sqrt{2}}\Big)^{2}=1$
$\dfrac{1}{2}+\dfrac{1}{2}=1$
$1=1$ True
Substituting the point $\Big(\dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{2}}\Big)$ into the equation makes it true. This point is on the graph of the equation.
$\Big(\dfrac{\sqrt{3}}{2},\dfrac{1}{2}\Big)$
$x^{2}+y^{2}=1$
$\Big(\dfrac{\sqrt{3}}{2}\Big)^{2}+\Big(\dfrac{1}{2}\Big)^{2}=1$
$\dfrac{3}{4}+\dfrac{1}{4}=1$
$1=1$ True
Substituting the point $\Big(\dfrac{\sqrt{3}}{2},\dfrac{1}{2}\Big)$ into the equation makes it true. This point is on the graph of the equation.