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 $A\left(2,0\right)\mathrm{and}B\left(-2,0\right)$. Let P(h, k) be a point such that $\angle APB={90}^{\circ }$.

Thus, $∆$APB is a right angled triangle.

$\therefore A{B}^2=A{P}^2+B{P}^2$

$$\begin{gathered} \therefore {\left(2+2\right)}^2+0={\left(h-2\right)}^2+k^2+{\left(h+2\right)}^2+k^2 \\ \Rightarrow 16=h^2+4-4h+k^2+h^2+4+4h+k^2 \\ \Rightarrow h^2+k^2=4 \end{gathered}$$

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