A circle passes through the intersection point of any two of the lines $y=x, y=2x$ and $y=15-0.5x$. What is the length of the chord that the circle cuts off on the line $x+y=16$?
The task can be solved arbitrarily with lines (unless there are parallel ones among them). However, in the given case, significant simplifications are possible compared to the generally applicable procedure.

![](https://cdn.mathpix.com/cropped/2024_05_02_e17187640ebeab5801c5g-1.jpg?height=781&width=645&top_left_y=260&top_left_x=729)

Notice that the second and third given lines are perpendicular to each other because the product of their slopes is \(2 \cdot (-0.5) = -1\). Thus, the three intersection points form a right triangle, and the center \(C\) of the circumscribed circle is at the midpoint of the hypotenuse, with the radius equal to half the hypotenuse. Therefore, it is sufficient to determine the two intersection points on the first line. We see that one of them is the origin, and the other is the point \((10, 10)\) from the first and third equations. From this, \(C(5, 5)\) and the radius is \(5\sqrt{2}\) units.

We can also skip calculating the intersection points of the fourth line because this line is perpendicular to the hypotenuse of the triangle, which is the diameter of the circle. The midpoint \(F\) of the chord is found from the first and fourth equations at the point \(F(8, 8)\). The distance of the chord from the center is \(CF = 3\sqrt{2}\), so the length of half the chord is: \(h/2 = \sqrt{r^2 - CF^2} = 4\sqrt{2}\), thus \(h = 8\sqrt{2}\) length units.