#### Question

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 difference of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 meters. Is it the possible to do so? If yes, at what distances from the two gates should the pole be erected?

#### Solution

Let *P *be the required location on the boundary of a circular park such that its distance from gate *B *is x metre that is *BP* x metres.

Then, *AP = x + 7*

In the right triangle *ABP *we have by using Pythagoras theorem

AP^{2} + BP^{2} = AB^{2}

(x + 7)^{2} + x^{2} = (13)^{2}

x^{2} + 14x + 49 + x^{2} = 169

2x^{2} + 14x + 49 - 169 = 0

2x^{2} + 14x - 120 = 0

2(x^{2} + 7x - 60) = 0

x^{2} + 7x - 60 = 0

x^{2} + 12x - 5x - 60 = 0

x(x + 12) - 5(x - 12) = 0

(x + 12)(x - 5) = 0

x + 12 = 0

x = -12

Or

x - 5 = 0

x = 5

But the side of right triangle can never be negative

Therefore, x = 5

Hence, *P *is at a distance of 5 metres from the gate *B*.

⇒ BP = 5m

Now, AP = (BP + 7)m = (5 + 7)m = 12 m

∴ The pole has to be erected at a distance 5 mtrs from the gate B and 12 m from the gate A.