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Geometry and Trigonometry / Circles Difficulty: Hard

A circle in the xy-plane has its center at $left parenthesis negative 5 comma 2 right parenthesis$ and has a radius of $9$ . An equation of this circle is $x^{2} + y^{2} + a x + b y + c = 0$ , where $a$ , $b$ , and $c$ are constants. What is the value of $c$ ?

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Explanation

The correct answer is $negative 52$ . 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 $left parenthesis negative 5 comma 2 right parenthesis$ and has a radius of $9$ . Substituting $negative 5$ 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 $x^{2} + y^{2} + a x + b y + c = 0$ , where $a$ , $b$ , and $c$ are constants. Therefore, ${(x + 5)}^{2} + {(y - 2)}^{2} = 81$ can be rewritten in the form $x^{2} + y^{2} + a x + b y + c = 0$ . The equation ${(x + 5)}^{2} + {(y - 2)}^{2} = 81$ , or $(x + 5) (x + 5) + (y - 2) (y - 2) = 81$ , can be rewritten as $x^{2} + 5 x + 5 x + 25 + y^{2} - 2 y - 2 y + 4 = 81$ . Combining like terms on the left-hand side of this equation yields $x^{2} + y^{2} + 10 x - 4 y + 29 = 81$ . Subtracting $81$ from both sides of this equation yields $x^{2} + y^{2} + 10 x - 4 y - 52 = 0$ , which is equivalent to $x^{2} + y^{2} + 10 x + (- 4) y + (- 52) = 0$ . This equation is in the form $x^{2} + y^{2} + a x + b y + c = 0$ . Therefore, the value of $c$ is $negative 52$ .