SAT Question 27f1db42: Circles answer and explanation with Desmos solution

An equation of a circle in the xy-plane can be written as $(x-h{)}^2+(y-k{)}^2=r^2$, where the center of the circle is $(h,k)$ and the radius of the circle is $r$. It’s given that in the xy-plane, circle $F$ is defined by the equation $(x+8{)}^2+(y+8{)}^2=25$, which can be rewritten as $(x-(-8){)}^2+(y-(-8){)}^2=5^2$. It follows that $h=-8$, $k=-8$, and $r=5$. Therefore, the center of circle $F$ is $(-8,-8)$ and the radius of circle $F$ is $5$. It’s given that circle $G$ has the same center as circle $F$, and the radius of circle $G$ is $2$ units greater than the radius of circle $F$. Therefore, the center of circle $G$ is $(-8,-8)$ and the radius of circle $G$ is $5+2$, or $7$. Substituting $-8$ for $h$, $-8$ for $k$, and $7$ for $r$ into the equation $(x-h{)}^2+(y-k{)}^2=r^2$ yields $(x+8{)}^2+(y+8{)}^2=7^2$, which is equivalent to $(x+8{)}^2+(y+8{)}^2=49$. It follows that circle $G$ is defined by the equation $(x+8{)}^2+(y+8{)}^2=49$; therefore, the value of $p$ is $49$.