Given, equation of circle is
$$
x^{2}+y^{2}=2 \quad \text{...(i)}
$$
Taking derivative w.r.t. ' $t$ ' on both sides.
$$
2 x \frac{d x}{d t}+2 y \frac{d y}{d t}=0 \Rightarrow x \frac{d x}{d t}+y \frac{d y}{d t}=0
$$
If abscissa and ordinate increase at the same rate, we have
$$
\frac{d x}{d t}=\frac{d y}{d t}
$$
$$
x \frac{d x}{d t}+y \frac{d x}{d t}=0 \Rightarrow \frac{d x}{d t}(x+y)=0
$$
Since,
$$
\Rightarrow \quad x+y=0 \Rightarrow x=-y \quad \text{...(ii)}
$$
Solving Eqs. (i) and (ii), we get
$$
x^{2}+(-x)^{2}=2 \Rightarrow x=\pm 1
$$
For $x=1, y=-1$ and $x=-1, y=1$
Required point are $(1,-1)$ and $(-1,1)$.