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Geometry and Trigonometry / Circles Difficulty: Hard

The graph of $x^{2} + x + y^{2} + y = \frac{199}{2}$ in the xy-plane is a circle. What is the length of the circle’s radius?

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Explanation

The correct answer is $10$ . It's given that the graph of $x^{2} + x + y^{2} + y = \frac{199}{2}$ in the xy-plane is a circle. The equation of a circle in the xy-plane can be written in the form ${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$ , where the coordinates of the center of the circle are $(h, k)$ and the length of the radius of the circle is $r$ . The term ${(x - h)}^{2}$ in this equation can be obtained by adding the square of half the coefficient of $x$ to both sides of the given equation to complete the square. The coefficient of $x$ is $1$ . Half the coefficient of $x$ is $\frac{1}{2}$ . The square of half the coefficient of $x$ is $\frac{1}{4}$ . Adding $\frac{1}{4}$ to each side of $(x^{2} + x) + (y^{2} + y) = \frac{199}{2}$ yields $(x^{2} + x + \frac{1}{4}) + (y^{2} + y) = \frac{199}{2} + \frac{1}{4}$ , or ${(x + \frac{1}{2})}^{2} + (y^{2} + y) = \frac{199}{2} + \frac{1}{4}$ . Similarly, the term ${(y - k)}^{2}$ can be obtained by adding the square of half the coefficient of $y$ to both sides of this equation, which yields ${(x + \frac{1}{2})}^{2} + (y^{2} + y + \frac{1}{4}) = \frac{199}{2} + \frac{1}{4} + \frac{1}{4}$ , or ${(x + \frac{1}{2})}^{2} + {(y + \frac{1}{2})}^{2} = \frac{199}{2} + \frac{1}{4} + \frac{1}{4}$ . This equation is equivalent to ${(x + \frac{1}{2})}^{2} + {(y + \frac{1}{2})}^{2} = 100$ , or ${(x + \frac{1}{2})}^{2} + {(y + \frac{1}{2})}^{2} = 10^{2}$ . Therefore, the length of the circle's radius is $10$ .