SAT Question 196e8e6e: Circles answer and explanation with Desmos solution

Choice A is correct. It's given that points $A$ and $B$ lie on the circle with center $C$. Therefore, $\overline{AC}$ and $\overline{BC}$ are both radii of the circle. Since all radii of a circle are congruent, $\overline{AC}$ is congruent to $\overline{BC}$. The length of $\overline{AC}$, or the distance from point $A$ to point $C$, can be found using the distance formula, which gives the distance between two points, $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$, as $\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2}$. Substituting the given coordinates of point $A$, $\left(h+1,k+\sqrt{102}\right)$, for $\left(x_1,y_1\right)$ and the given coordinates of point $C$, $\left(h,k\right)$, for $\left(x_2,y_2\right)$ in the distance formula yields $\sqrt{{\left(h+1-h\right)}^2+{\left(k+\sqrt{102}-k\right)}^2}$, or $\sqrt{1^2+{\left(\sqrt{102}\right)}^2}$, which is equivalent to $\sqrt{1+102}$, or $\sqrt{103}$. Therefore, the length of $\overline{AC}$ is $\sqrt{103}$ and the length of $\overline{BC}$ is $\sqrt{103}$. It's given that angle $ACB$ is a right angle. Therefore, triangle $ACB$ is a right triangle with legs $\overline{AC}$ and $\overline{BC}$ and hypotenuse $\overline{AB}$. By the Pythagorean theorem, if a right triangle has a hypotenuse with length $c$ and legs with lengths $a$ and $b$, then $a^2+b^2=c^2$. Substituting $\sqrt{103}$ for $a$ and $b$ in this equation yields ${\left(\sqrt{103}\right)}^2+{\left(\sqrt{103}\right)}^2=c^2$, or $103+103=c^2$, which is equivalent to $206=c^2$. Taking the positive square root of both sides of this equation yields $\sqrt{206}=c$. Therefore, the length of $\overline{AB}$ is $\sqrt{206}$.