Question

A pole has to be erected at a point on the boundary of a circular park of diameter 17 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 1

Hint: The most important thing that is to be known in this question is that as the two gates A and B are diametrically opposite in the circular park and the pole has to be erected at the boundary, therefore, A and B will subtend a right angle at the point of the erection of the pole. Another important formula of Pythagoras theorem is used \[{{A}^{2}}={{B}^{2}}+{{C}^{2}}\] to solve this question.

Complete step-by-step answer:
As mentioned in the question, we have to find the possibility as well as the point at which the pole is to be made erect.

Now, let us take the distance between the pole and gate A be x and the distance between the pole and the gate B be y.
Using Pythagoras theorem, we get that
\[\begin{align}
  & {{x}^{2}}+{{y}^{2}}={{17}^{2}} \\
 & {{x}^{2}}+{{y}^{2}}=289\ \ \ \ \ ...(a) \\
\end{align}\]
Now, the difference between the pole’s distances from gate A and B is as follows
\[\left( x-y \right)=7\]
Using the above equation in equation (a), we get
\[\begin{align}
  & {{x}^{2}}+{{y}^{2}}=289 \\
 & {{\left( y+7 \right)}^{2}}+{{y}^{2}}=289 \\
 & 2{{y}^{2}}+14y+49=289 \\
 & 2{{y}^{2}}+14y=240 \\
 & {{y}^{2}}+7y-120=0 \\
\end{align}\]
Now, the discriminant of this equation is as follows
\[\begin{align}
  & D={{7}^{2}}-(-120)4 \\
 & D=49+480 \\
 & D=529 \\
\end{align}\]
As the discriminant of the equation is positive, hence, such a case is possible.
Now, on solving the quadratic equation
$\Rightarrow{{y}^{2}}+7y-120=0$
$\Rightarrow{{y}^{2}}-8y+15y-120=0$
$\Rightarrow y(y-8)+15(y-8)$
$\Rightarrow {(y-8) { \& } (y+15)}$
Therefore $y$ = $8$ and $y$ = $-15$
Now, distance cannot be negative, therefore, the value is 8.
$y=PB=8m$
The difference of pole’s distance from gate A and gate B given
i.e $x-y=7$
$x=7+y$
Substituting the value of $y$ ,we get
$x=7+8$
$x=15m$ which is $PA$
Hence, the distance between pole and gate A is 15 and gate B is 8.

Note: The students can make an error if they don’t know the properties of a circle that is given in the hint that is to be known in this question is that as the two gates A and B are diametrically opposite in the circular park and the pole has to be erected at the boundary, therefore, A and B will subtend a right angle at the point of the erection of the pole.