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Geometry and Trigonometry Difficulty: Medium

$x^{2} + 58 x + y^{2} = 0$ $x^{2} + 58 x + y^{2} = 0$

In the xy-plane, the graph of the given equation is a circle. What are the coordinates $(x, y)$ $(x, y)$ of the center of the circle?

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Explanation

Choice D is correct. It’s given that in the xy-plane, the graph of $x^{2} + 58 x + y^{2} = 0$ $x^{2} + 58 x + y^{2} = 0$ is a circle. The equation of a circle in the xy-plane can be written as ${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$ ${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$ , where the coordinates of the center of the circle are $(h, k)$ $(h, k)$ and the radius of the circle is $r$ $r$ . By completing the square, the equation $x^{2} + 58 x + y^{2} = 0$ $x^{2} + 58 x + y^{2} = 0$ can be rewritten as $(x^{2} + 58 x + {(\frac{58}{2})}^{2}) + y^{2} = 0 + {(\frac{58}{2})}^{2}$ $(x^{2} + 58 x + {(\frac{58}{2})}^{2}) + y^{2} = 0 + {(\frac{58}{2})}^{2}$ , or $(x^{2} + 58 x + 841) + y^{2} = 841$ $(x^{2} + 58 x + 841) + y^{2} = 841$ . This equation is equivalent to ${(x + 29)}^{2} + y^{2} = 841$ ${(x + 29)}^{2} + y^{2} = 841$ , or ${(x - (- 29))}^{2} + {(y - 0)}^{2} = 841$ ${(x - (- 29))}^{2} + {(y - 0)}^{2} = 841$ . Therefore, $h$ $h$ is $- 29$ $negative 29$ and $k$ $k$ is $0$ $0$ , and the coordinates $(x, y)$ $(x, y)$ of the center of the circle are $(- 29, 0)$ $left parenthesis negative 29 comma 0 right parenthesis$ .

Choice A is incorrect and may result from conceptual or calculation errors.

Choice B is incorrect and may result from conceptual or calculation errors.

Choice C is incorrect and may result from conceptual or calculation errors.