Problem 1470

The two equations can be rewritten as
$$\begin{aligned} (x - 24)^2 + \left(y - \frac{29}{2}\right)^2 &= \frac{289}{4} \\ (x - 24)\left(y - \frac{29}{2}\right) &= -30. \end{aligned}$$
By putting $X = x - 24$ and $Y = y - \frac{29}{2}$, we obtain
$$X^2 + Y^2 = \frac{289}{4} \tag{1}$$
$$XY = -30. \tag{2}$$
From (2) we have $Y = -30/X$. Plug this into (1) to get $X^2 + \frac{900}{X^2} = \frac{289}{4}$, which simplifies to
$$X^4 - \frac{289}{4}X^2 + 900 = 0.$$
We can now solve for $X^2$, using the quadratic formula:
$$\begin{aligned} X^2 &= \frac{1}{2} \left( \frac{289}{4} \pm \sqrt{\left(\frac{289}{4}\right)^2 - 60^2} \right) \\ &= \frac{1}{2} \left( \frac{289}{4} \pm \sqrt{\left(\frac{289}{4} - 60\right) \left(\frac{289}{4} + 60\right)} \right) \\ &= \frac{1}{2} \left( \frac{289}{4} \pm \sqrt{\frac{49}{4} \cdot \frac{529}{4}} \right) \\ &= \frac{1}{2} \left( \frac{289 \pm 161}{4} \right), \end{aligned}$$
so that $X^2 = \frac{225}{4}$ or $X^2 = 16$. Thus, $X = \pm\frac{15}{2}$ or $X = \pm 4$, and correspondingly, $Y = \mp 4$ or $Y = \mp\frac{15}{2}$. Using $x = X + 24$ and $y = Y + \frac{29}{2}$, we find that $(x, y)$ solves the original set of equations if and only if $(x, y) \in \{(28, 7), (20, 22), (\frac{33}{2}, \frac{37}{2}), (\frac{63}{2}, \frac{21}{2})\}$.

Proceed as in Solution 1 up to the two equations
$$X^2 + Y^2 = \frac{289}{4} \tag{1}$$
$$XY = -30. \tag{2}$$
By either adding $2XY = -60$ to (1), or subtracting it from (1), we get, respectively,
$$\begin{aligned} X^2 + 2XY + Y^2 &= (X + Y)^2 = \frac{49}{4} = \left(\frac{7}{2}\right)^2 \\ X^2 - 2XY + Y^2 &= (X - Y)^2 = \frac{529}{4} = \left(\frac{23}{2}\right)^2. \end{aligned}$$
From this, we have to solve four systems of equations:
$$\begin{aligned} X + Y &= \pm\frac{7}{2} \\ X - Y &= \pm\frac{23}{2} \end{aligned}$$