Question

Find the locus of a point such that the line segments with end points (2, 0) and (−2, 0) subtend a right angle at that point.

 

Answer 1

Let the given points be \[A\left( 2, 0 \right)\text{ and }B\left( - 2, 0 \right)\]. Let P(h, k) be a point such that

\[\angle APB = {90}^\circ\]
Thus,
∆ APB is a right angled triangle.
\[\therefore A B^2 = A P^2 + B P^2\]

\[\therefore \left( 2 + 2 \right)^2 + 0 = \left( h - 2 \right)^2 + k^2 + \left( h + 2 \right)^2 + k^2 \]
\[ \Rightarrow 16 = h^2 + 4 - 4h + k^2 + h^2 + 4 + 4h + k^2 \]
\[ \Rightarrow h^2 + k^2 = 4\]

Hence, the locus of (h, k) is x²+ y² = 4.