sat-dog

The correct answer is $-52$. The equation of a circle in the xy-plane with its center at $\left(h,k\right)$ and a radius of $r$ can be written in the form ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$. It's given that a circle in the xy-plane has its center at $\left(-5,2\right)$ and has a radius of $9$. Substituting $-5$ for $h$, $2$ for $k$, and $9$ for $r$ in the equation ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ yields ${\left(x-\left(-5\right)\right)}^2+{\left(y-2\right)}^2=9^2$, or ${\left(x+5\right)}^2+{\left(y-2\right)}^2=81$. 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, ${\left(x+5\right)}^2+{\left(y-2\right)}^2=81$ can be rewritten in the form $x^2+y^2+ax+by+c=0$. The equation ${\left(x+5\right)}^2+{\left(y-2\right)}^2=81$, or $\left(x+5\right)\left(x+5\right)+\left(y-2\right)\left(y-2\right)=81$, 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 $x^2+y^2+10x+\left(-4\right)y+\left(-52\right)=0$. This equation is in the form $x^2+y^2+ax+by+c=0$. Therefore, the value of $c$ is $-52$.