Answer
a) No
b) Yes
Work Step by Step
The points are on the graph when $x^2+y^2=1$
Check the points:
a) $x^2+y^2=(\dfrac{1}{2})^2+(\dfrac{1}{2})^2=\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{2}\ne1$
Therefore, this point is not on the graph.
b) $x^2+y^2=(\dfrac{1}{2})^2+(\dfrac{\sqrt3}{2})^2=\dfrac{1}{4}+\dfrac{3}{4}=\dfrac{4}{4}=1$
Therefore, this point is on the graph.
