The point set given by the equation $y\left(x^{2}+y^{2}\right)-x\left(x^{2}+y^{2}\right)-y+x=0$ which point is closest to the point $P(3 ; 4)$?
Transform the left side of the equation into a product:

$$
y\left(x^{2}+y^{2}\right)-x\left(x^{2}+y^{2}\right)-y+x=(y-x)\left(x^{2}+y^{2}-1\right)=0
$$

The first factor is the equation of the angle bisector passing through the origin in the first quadrant, and the second factor is the equation of a unit circle centered at the origin. The equation is satisfied by the points in the union of the two sets.

![](https://cdn.mathpix.com/cropped/2024_05_02_14204fc1a00f550fe0f6g-1.jpg?height=501&width=521&top_left_y=373&top_left_x=802)

The distance from point $P$ to the circle is the segment determined by the intersection of the line connecting $P$ and $P-\mathrm{t}$ with the origin and the circle at point $M_{1}$. This is the shortest segment connecting $P$ to the circumference of the circle. The distance between $P$ and $M_{1}$ is $d_{1}=\sqrt{3^{2}+4^{2}}-1=4$ units.

Among the segments drawn from a point to the points of a line not passing through it, the shortest is the perpendicular segment. Let's calculate its length. The equation of the perpendicular from $P$ to the line $y=x$ is:

$$
y-4=-(x-3), \quad \text { rearranging } \quad y=-x+7
$$

The coordinates of the intersection of the two lines are obtained from their equations, giving $M_{2}(3.5 ; 3.5)$, and finally the distance between the two points, $P$ and $M_{2}$, is

$$
d_{1}=\sqrt{(3-3.5)^{2}+(4-3.5)^{2}}=\frac{\sqrt{2}}{2} \approx 0.7071
$$

which is smaller than $d_{1}=4$.

Therefore, the point $M_{2}(3.5 ; 3.5)$ of the set defined by the equation is the closest to $P$. Adam Vanya (Esztergom, Dobó K. Gymnasium, 11th grade)