Let the orthogonal projection of the origin of the Cartesian coordinate system onto the line

$$
\frac{x}{a}+\frac{y}{b}=1
$$

be $P$. What is the geometric locus of the points $P$ if the line moves in such a way that $\frac{1}{a^{2}}+\frac{1}{b^{2}}$ remains constant?
I. solution: Calculate the coordinates of the orthogonal projection of the starting point, the point $P$. The direction coefficient of the line

$$
\frac{x}{a}+\frac{y}{b}=1
$$

is $-\frac{b}{a}$, so the equation of the line $O P$ perpendicular to it is

$$
y=\frac{a}{b} x
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_72b755367d554c807e1eg-1.jpg?height=488&width=643&top_left_y=515&top_left_x=730)

The point $P$ is the intersection of (1) and (2); its coordinates are

$$
\begin{aligned}
x & =\frac{a b^{2}}{a^{2}+b^{2}} \\
y & =\frac{a^{2} b}{a^{2}+b^{2}}
\end{aligned}
$$

According to the problem, we make the following assumption for $a$ and $b$:

$$
\frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{1}{c^{2}}, \text { where } c \text { is a constant. }
$$

From (5),

$$
c^{2}=\frac{1}{\frac{1}{a^{2}}+\frac{1}{b^{2}}}=\frac{a^{2} b^{2}}{a^{2}+b^{2}}
$$

The equation of the geometric locus we are looking for is obtained by eliminating $a$ and $b$ from equations (3), (4), and (5), which hold for every point of the geometric locus.

The sum of the squares of (3) and (4), taking (6) into account,

$$
x^{2}+y^{2}=\frac{a^{2} b^{4}+a^{4} b^{2}}{\left(a^{2}+b^{2}\right)^{2}}=\frac{a^{2} b^{2}\left(a^{2}+b^{2}\right)}{\left(a^{2}+b^{2}\right)^{2}}=\frac{a^{2} b^{2}}{a^{2}+b^{2}}=c^{2}
$$

Thus, the geometric locus we are looking for is a circle centered at the origin with radius

$$
c=\frac{a b}{\sqrt{a^{2}+b^{2}}}
$$

László Perneczky (Kaposvár, Táncsics g. III. o. t.)

II. solution: Calculate the length of the segment $O P$. Expressing twice the area of the right triangle with legs $a, b$ in two ways,

$$
2 t=a b=O P \cdot \sqrt{a^{2}+b^{2}}
$$

from which

$$
O P=\frac{a b}{\sqrt{a^{2}+b^{2}}}=c \text { (constant) }
$$

which means that the geometric locus of the points $P$ is a circle centered at $O$ with radius $c=\frac{a b}{\sqrt{a^{2}+b^{2}}}$.

III. solution: Let the coordinate system coincide with a $\Sigma\left(90^{\circ}\right)$ system. In problem 675, we proved that the equation of the geometric locus of the points adjoint to the tangents of a circle with radius $r$ centered at $O$ is

$$
\frac{1}{x^{2}}+\frac{1}{y^{2}}=\frac{1}{r^{2}}
$$

and according to the interpretation of the $\Sigma\left(90^{\circ}\right)$ system, the coordinates of the point are equal to the intercepts of the adjoint line: $x=a, y=b$. Thus, all lines whose intercepts satisfy the condition (7) are tangents to a circle with radius

$$
r=\sqrt{\frac{x^{2} y^{2}}{x^{2}+y^{2}}}=\frac{a b}{\sqrt{a^{2}+b^{2}}}
$$

and the feet of the perpendiculars from $O$ to the tangents are points on this circle.