SAT Question 44d67c6c: Circles answer and explanation with Desmos solution

It’s given that $x^2+y^2-10x-8y-14n=0$ represents circle A in the xy-plane, where $n$ is a constant. 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 of the circle and $r$ is its radius. By completing the square, the equation $x^2+y^2-10x-8y-14n=0$ can be rewritten as $(x^2-10x+(\frac{-10}{2}{)}^2)+(y^2-8y+(\frac{-8}{2}{)}^2)=14n+(\frac{-10}{2}{)}^2+(\frac{-8}{2}{)}^2$. This is equivalent to $(x^2-10x+25)+(y^2-8y+16)=14n+25+16$, which can be written as $(x-5{)}^2+(y-4{)}^2=14n+41$. It follows that $(5,4)$ is the center of circle A and $\sqrt{14n+41}$ is its radius. It’s given that circle B has the same center as circle A. Therefore, the center of circle B is $(5,4)$. It's also given that circle B has twice the diameter as circle A. Therefore, circle B has twice the radius as circle A. Since the radius of circle A is $\sqrt{14n+41}$, it follows that the radius of circle B is $2\sqrt{14n+41}$. Substituting $5$ for $h$, $4$ for $k$, and $2\sqrt{14n+41}$ for $r$ in the equation $(x-h{)}^2+(y-k{)}^2=r^2$ yields $(x-5{)}^2+(y-4{)}^2={\left(2\sqrt{14n+41}\right)}^2$, or $(x-5{)}^2+(y-4{)}^2=4(14n+41)$, which represents an equation for circle B. It’s given that the point $(11,8)$ lies on circle B. Substituting $11$ for $x$ and $8$ for $y$ in the equation $(x-5{)}^2+(y-4{)}^2=4(14n+41)$ yields $(11-5{)}^2+(8-4{)}^2=4(14n+41)$, which gives $6^2+4^2=4(14n+41)$, or $52=4(14n+41)$. Dividing both sides of this equation by $4$ yields $13=14n+41$. Subtracting $41$ from both sides of this equation yields $-28=14n$. Dividing both sides of this equation by $14$ yields $-2=n$. Therefore, the value of $n$ is $-2$.