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

(A) Let $P$ be the required location of the pole on the boundary of the circular park.
Let the distance of the pole from gate $B$ be $x \, m$,i.e.,$BP = x \, m$.
The difference of the distances of the pole from the two gates is $AP - BP = 7 \, m$.
Therefore,$AP = (x + 7) \, m$.
Since $AB$ is the diameter of the circular park,$AB = 13 \, m$.
In a circle,the angle subtended by the diameter at any point on the boundary is $90^{\circ}$. Thus,$\angle APB = 90^{\circ}$.
By Pythagoras theorem in $\triangle APB$,we have $AP^2 + BP^2 = AB^2$.
Substituting the values,$(x + 7)^2 + x^2 = 13^2$.
Expanding the equation: $x^2 + 14x + 49 + x^2 = 169$.
$2x^2 + 14x - 120 = 0$.
Dividing by $2$,we get the quadratic equation: $x^2 + 7x - 60 = 0$.
To check if this is possible,we calculate the discriminant $D = b^2 - 4ac = 7^2 - 4(1)(-60) = 49 + 240 = 289$.
Since $D > 0$,the equation has two distinct real roots,so it is possible to erect the pole.
Using the quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$,we get $x = \frac{-7 \pm \sqrt{289}}{2} = \frac{-7 \pm 17}{2}$.
This gives $x = \frac{10}{2} = 5$ or $x = \frac{-24}{2} = -12$.
Since distance cannot be negative,we take $x = 5$.
Thus,the pole should be erected at a distance of $5 \, m$ from gate $B$ and $12 \, m$ from gate $A$.