# 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 metres. Is it possible to do so? If yes, at what distances from the two gates should the pole be erected?

Let P be the position of the pole and A and B be the opposite fixed gates. PA = ‘a’ mts

PB = ‘b’ mts

PA – PB = 7 m

a – b = 7

a = 7 + b .........(1)

In Δ PAB,

AB2 = AP2 + BP2

(17) = (a)2 + (b)2

a2 + b2 = 289

Putting the value of a = 7 + b in the above, (7 + b)2 + b2 = 289

49 + 14b + 2b2 = 289

2b2 + 14b + 49 – 289 = 0

2b2 + 14b – 240 = 0

Dividing the above by 2, we get.

b2 + 7b – 120 = 0

b2 + 15b – 8b – 120 = 0

b(b + 15) – 8(b + 15) = 0

(b – 8) (b + 15) = 0

b = 8 or b = –15

Since this value cannot be negative, so b = 8 is the correct value.

Yes it is possible to erect a pole.

Putting b = 8 in (1), we get.

a = 7 + 8

a = 15 m

Hence PA = 15 m and PB = 8 m

So, the distance from the gate A to pole is 15 m and from gate B to the pole is 8 m.

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