Answer
a. $\large (\frac{3}{4},\frac{\sqrt{7}}{4})$
Work Step by Step
The equation of the unit circle is
$$x^2+y^2=1$$
For point $a$ $\large (\frac{3}{4},\frac{\sqrt{7}}{4})$
$$x^2+y^2 = \left(\frac{3}{4}\right)^2 + \left(\frac{\sqrt{7}}{4}\right)^2 = 1$$
$\therefore$ Point $a$ lies on the unit circle.
For point $b$ $(\sqrt{2},\sqrt{2})$
$$x^2+y^2 = (\sqrt{2})^2 + (\sqrt{2})^2= 4$$
$\therefore$ Point $b$ doesn't lie on the unit circle.
For point $c$ $ (1,\sqrt{3})$
$$x^2+y^2 = (1)^2+(\sqrt{3})^2 = 4$$
$\therefore$ Point $c$ doesn't lie on the unit circle.
For point $d$ $\large(\frac{1}{3}, \frac{\sqrt{2}}{3})$
$$x^2+y^2 = \left(\frac{1}{3}\right)^2 +\left (\frac{\sqrt{2}}{3}\right)^2 = \frac{1}{3}$$
$\therefore$ Point $d$ doesn't lie on the unit circle.
