Circles SAT question

The correct answer is $-52$. The equation of a circle in the xy-plane with its center at $(h,k)$ and a radius of $r$ can be written in the form ${(x-h)}^2+{(y-k)}^2=r^2$. It's given that a circle in the xy-plane has its center at $(-5,2)$ and has a radius of $9$. Substituting $-5$ for $h$, $2$ for $k$, and $9$ for $r$ in the equation ${(x-h)}^2+{(y-k)}^2=r^2$ yields ${(x-(-5))}^2+{(y-2)}^2=9^2$, or ${(x+5)}^2+{(y-2)}^2=81$. It's also given that an equation of this circle is $x^2+y^2+ax+by+c=0$, where $a$, $b$, and $c$ are constants. Therefore, ${(x+5)}^2+{(y-2)}^2=81$ can be rewritten in the form $x^2+y^2+ax+by+c=0$. The equation ${(x+5)}^2+{(y-2)}^2=81$, or $(x+5)(x+5)+(y-2)(y-2)=81$, can be rewritten as $x^2+5x+5x+25+y^2-2y-2y+4=81$. Combining like terms on the left-hand side of this equation yields $x^2+y^2+10x-4y+29=81$. Subtracting $81$ from both sides of this equation yields $x^2+y^2+10x-4y-52=0$, which is equivalent to $x^2+y^2+10x+(-4)y+(-52)=0$. This equation is in the form $x^2+y^2+ax+by+c=0$. Therefore, the value of $c$ is $-52$.