SAT Question 408b39ea: Circles answer and explanation with Desmos solution

Choice A is correct. It’s given that the graph of the equation $(x-3{)}^2+(y+16{)}^2=289$ is a circle in the xy-plane. For a circle in the form $(x-h{)}^2+(y-k{)}^2=r^2$, its center is $(h,k)$ and its radius is $r$. It follows that $r^2=289$. Taking the positive square root of both sides of this equation yields $r=17$. Therefore, the center of this circle is $(3,-16)$ and the radius is $17$. It follows that the x-coordinates of the points that lie on the circle must be within $17$ units of the x-coordinate of the center of the circle. Therefore, the least possible x-coordinate of a point that lies on the circle is $3-17$, or $-14$, and the greatest possible x-coordinate of a point that lies on the circle is $3+17$, or $20$. So for any point $(q,s)$ to lie on the circle, it must be true that $-14\le q\le 20$. Since $-17$ does not satisfy this inequality, it is NOT a possible value of $q$.