A circular park has a diameter of 25 m. Two gates A and B are at the two ends of a diameter. A water tap is to be fixed on the boundary of the park such that the difference of its distances from the two gates is 17 m. Is it possible? If yes, find the…

Answer: Yes. The tap is 7 m from one gate and 24 m from the other.

Gates A and B are the ends of a diameter, and tap T is on the circle, so \(\angle ATB = 90^\circ\) (angle inscribed in a semicircle).

Let TA = x m and TB = (x + 17) m. By Pythagoras theorem, \(x^2 + (x + 17)^2 = 25^2\).

\(2x^2 + 34x + 289 = 625\) \(\therefore 2x^2 + 34x - 336 = 0\) \(\therefore x^2 + 17x - 168 = 0\).

\((x + 24)(x - 7) = 0\) \(\therefore x = 7\) (distance cannot be −24).

So the tap can be fixed: 7 m from one gate and 7 + 17 = 24 m from the other. Check: \(49 + 576 = 625 = 25^2\). (There are four such points on the boundary: two on each side of diameter AB.)