The coordinates of the vertices of a certain triangle are: $(a, 0)(b, 0)(0, c)$. Let's determine the equation of the Feuerbach circle of this triangle!
The Feuerbach circle passes through the midpoints of the sides; their coordinates are:

$$
\left(\frac{a+b}{2}, 0\right), \quad\left(\frac{a}{2}, \frac{c}{2}\right), \quad\left(\frac{b}{2}, \frac{c}{2}\right) \ldots
$$

The equation of the circle is:

$$
x^{2}+y^{2}+A x+B y+C=0 \ldots
$$

Here, the coefficients $A, B, C$ need to be determined. Since the circle 2) passes through the 3 points listed in 1), their coordinates satisfy the equation 2), and thus we get 3 linear equations for the calculation of $A, B, C$:

$$
\begin{aligned}
\left(\frac{a+b}{2}\right)^{2}+\left(\frac{a+b}{2}\right) A+C & =0 \ldots \\
\left(\frac{a}{2}\right)^{2}+\left(\frac{c}{2}\right)^{2}+\frac{a}{2} A+\frac{c}{2} B+C & =0 \ldots \\
\left(\frac{b}{2}\right)^{2}+\left(\frac{c}{2}\right)^{2}+\frac{b}{2} A+\frac{c}{2} B+C & =0 \ldots
\end{aligned}
$$

By subtracting the corresponding terms of 4) and 5):

$$
\left(\frac{a}{2}\right)^{2}-\left(\frac{b}{2}\right)^{2}+\frac{a-b}{2} A=0 \text { and thus } A=\frac{a+b}{2} \ldots
$$

Substituting the value of $A$ into 3):

$$
C=0 \ldots
$$

Substituting the values of $A$ and $C$ into, for example, 4):

$$
\frac{a^{2}+c^{2}}{4}-\frac{a(a+b)}{4}+\frac{c}{2} B=0 \text { and thus } B=\frac{a b-c^{2}}{2 c} \ldots
$$

Therefore, the equation of the Feuerbach circle, in this case, is:

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

This circle passes through the origin. Indeed, the origin is one of the feet of the altitudes: the Feuerbach circle passes through the feet of the altitudes as well.

Rudolf Szkitsák and János Szabó (Royal Catholic Grammar School VIII. o. Bp. II.)