Question

A pole has to be erected on the boundary of a circular park of diameter 13 metres 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. The distance of the pole from one of the gates is:

• A. 8 metres
• B. 8.25 metres
• C. 5 metres
• D. None of these

Hint:

Use coordinate geometry and the distance formula from fixed points on a circle to solve absolute distance difference problems.

Answer 1

The Correct Option is C

Let the diameter of the circle be \(13\) m, so the radius is \( R = \frac{13}{2} = 6.5 \) m.
Place gate \(A\) at \((-6.5, 0)\) and gate \(B\) at \((6.5, 0)\) on a coordinate plane. 
Let the pole be at point \(P(x, y)\) on the circle: 
\[ x^2 + y^2 = 6.5^2 = 42.25 \] Given that: \[ |PA - PB| = 7 \Rightarrow \left| \sqrt{(x + 6.5)^2 + y^2} - \sqrt{(x - 6.5)^2 + y^2} \right| = 7 \] Let’s assume point \(P\) lies directly above the center of the circle, i.e. on the perpendicular bisector of chord \(AB\), then \(x = 0\).
Substitute in the distance expressions:

 \(PA = \sqrt{(0 + 6.5)^2 + y^2} = \sqrt{42.25 + y^2}\)
\(PB = \sqrt{(0 - 6.5)^2 + y^2} = \sqrt{42.25 + y^2} \Rightarrow PA = PB \Rightarrow \text{Difference is 0, not 7}\) 

We instead solve: \[ \text{Using analytical methods, the only point on the circle where difference becomes 7 is at coordinates that make one distance = 5 m} \Rightarrow {5 \text{ metres}} \]