A pole has to be erected at a point on the boundary of a circular park of diameter of 13 - Maths - Quadratic Equations

Question

A pole has to be erected at a point on the boundary of a
circular park of diameter of 13 metres in such a way that the
differences of it 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 be pole
be erected

Santosh Kumar · Accepted Answer

Let P be the position of the pole and A and B be the opposite fixed gates.

PA – PB = 7

⇒ a – b = 7

⇒ a = 7 + b ...(1)

In ΔPAB,

AB² = AP² + BP²

$\Rightarrow (13)^{2} = a^{2} + b^{2} \Rightarrow a^{2} + b^{2} = 169 \Rightarrow (7 + b)^{2} + b^{2} = 169 [from (1)] \Rightarrow 49 + 2 (7) b + b^{2} + b^{2} = 169 \Rightarrow 49 + 14 b + 2 b^{2} = 169 \Rightarrow 2 b^{2} + 14 b + 49 - 169 = 0 \Rightarrow 2 b^{2} + 14 b - 120 = 0 \Rightarrow b^{2} + 7 b - 60 = 0$

$\Rightarrow b^{2} + 12 b - 5 b - 60 = 0 \Rightarrow b (b + 12) - 5 (b + 12) = 0 \Rightarrow (b - 5) (b + 12) = 0 \Rightarrow b = 5 or b = - 12$

As b cannot be negative, so b = 5

∴ a = 7 + 5 = 12

Hence, PA = 12m and PB = 5m

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