SAT Question 8493fd10: Circles answer and explanation with Desmos solution

Choice B is correct. It's given that the equation $x^2+x\sqrt{t}+y^2-y-65=0$, where $t$ is a constant, defines a circle in the xy-plane. The equation of a circle in the xy-plane can be written in the form $(x-h{)}^2+(y-k{)}^2=r^2$, where $(h,k)$ is the center and $r$ is the radius of the circle. Adding $65$ to both sides of the given equation yields $x^2+x\sqrt{t}+y^2-y=65$. By completing the square, this equation can be rewritten as $\left(x^2+x\sqrt{t}+{\left(\frac{\sqrt{t}}{2}\right)}^2\right)+\left(y^2-y+{\left(-\frac{1}{2}\right)}^2\right)=65+{\left(\frac{\sqrt{t}}{2}\right)}^2+{\left(-\frac{1}{2}\right)}^2$. This equation can be rewritten as $\left(x^2+x\sqrt{t}+\frac{t}{4}\right)+\left(y^2-y+\frac{1}{4}\right)=65+\frac{t}{4}+\frac{1}{4}$, or ${\left(x+\frac{\sqrt{t}}{2}\right)}^2+{\left(y-\frac{1}{2}\right)}^2=\frac{261}{4}+\frac{t}{4}$. Therefore, $r^2=\frac{261}{4}+\frac{t}{4}$. It's given that the radius of this circle is $\sqrt{83}$; therefore, $r=\sqrt{83}$. Squaring both sides of this equation yields $r^2=83$. Since $r^2=\frac{261}{4}+\frac{t}{4}$ and $r^2=83$, it follows that $\frac{261}{4}+\frac{t}{4}=83$. Multiplying both sides of this equation by $4$ yields $261+t=332$. Subtracting $261$ from both sides of this equation yields $t=71$. Therefore, the value of $t$ is $71$.