A pole has to be erected at a point on the boundary of a circular park of diameter 13 metres in such a way that the differences of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 metres. Is it possible to do so? If yes, at what distances from the two gates should the pole be erected?
Step-by-Step Solution
Key Concept: Use Thales' theorem: the angle subtended by a diameter at any point on the circle is a right angle. Hence triangle formed by the two gates and the required point is a right‑angled triangle with hypotenuse equal to the diameter. Apply the Pythagorean theorem together with the given difference of the two sides, leading to a quadratic equation.
1. Let the circular park have centre O and diameter AB = 13 m. Points A and B are diametrically opposite gates.
2. Let P be the required point on the circumference. Denote PA = x m and PB = y m.
3. Since AB is a diameter, by Thales' theorem \(\angle APB = 90^{\circ}\). Therefore triangle APB is right‑angled with hypotenuse AB.
4. Apply the Pythagorean theorem:
$$x^{2}+y^{2}=AB^{2}=13^{2}=169.$$
5. The condition on the distances is \(|x-y|=7|\). Assume \(x>y\) (the other case gives the symmetric position), so
$$x-y=7 \quad\Rightarrow\quad x = y+7.$$
6. Substitute \(x = y+7\) into the Pythagorean relation:
$$(y+7)^{2}+y^{2}=169$$
$$y^{2}+14y+49+y^{2}=169$$
$$2y^{2}+14y-120=0$$
Divide by 2:
$$y^{2}+7y-60=0.$$
7. Factor the quadratic:
$$(y+12)(y-5)=0$$
Hence \(y=5\) (positive) or \(y=-12\) (reject as distance cannot be negative).
8. Find \(x\):
$$x = y+7 = 5+7 = 12\;\text{metres}.$$
9. Thus the required distances are 12 m from one gate and 5 m from the other. Because the figure is symmetric about the line AB, there are two possible positions of P on the circle, each giving the same pair of distances.
10. Since both distances are less than the diameter, the construction is feasible.
Correct Answer: Yes, it is possible. The pole can be placed so that its distances from the two gates are 12 m and 5 m (the two positions are symmetric with respect to the line AB).