Answer
$\left\{\begin{array}{l}
x^{2}+y^{2}\leq 2\\
x\geq 1
\end{array}\right.$
Work Step by Step
The circle with center $C(h,k)$ and radus $a$ has the general equation
$( x-h)^{2}+(y-k)^{2}=a^{2}.$
The center's coordinates satisfy the inequality
$( h-h)^{2}+(k-k)^{2}=0\lt a^{2}$,
so points on the circle or in the interior, satisfy
$( x-h)^{2}+(y-k)^{2}\leq a^{2}$
or, here
$x^{2}+y^{2}\leq 2$
The points to the right of the line $x=1$ (the vertical line that passes through the given point) satisfy the inequality: $\qquad x \gt 1.$
So, if we include the points on the line, $\qquad x\geq 1$
$\left\{\begin{array}{l}
x^{2}+y^{2}\leq 2\\
x\geq 1
\end{array}\right.$
