(Q2) A pole has to be erected at a point on the boundary of a circular park of diameter 13 meters in such a way that the differences of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 meters. Is it possible to do so? If yes, at what distances from the two gates should the pole be erected?

Answer

Let us first draw the diagram.

APB13m

Let P be the required location of the pole. Let the distance of the pole from the gate B be x m, i.e BP = x m. Now the difference of the distances of the pole from the two gates = AP - BP (or,BP - AP)= 7 m.
Therefore, AP =(x+7)m

Now, AB=13m, and since AB is a diameter,

∠APB=900

By pythagoras theorem

AP2 + PB2 = AB2

(x + 7)2+x2=2

x2 + 14x + 49 + x2 =

2x2+14x- = 0

So, the distance 'x' of the pole from gate B satisfies the equation

x2 + 7x - = 0

So, it would be possible to place the pole if this equattion has real roots. To see if this so or not, let us consider its discriminant.The discriminant is

b2 - 4ac = 72-4×1×(-)= > 0

So, the given quadratic equations has two real roots, and it is possible to erect the pole on the boundary of the park.

Solving the quadratic equation x2 + 7x - = 0, by the quadratic formula, we get
x =
-7±√
2
 =
-7±
2

Therfore, x = or -

Since x is the distance between the pole and the gate B, it must be positive.

Therefore, x = - will have to be ignored. So, x = .

Thus, the pole has to be erected on the boundary of the park at a distance of 5m from the gate B and 12m from the gate A.