The sum of the squares of the distances of a moving point from 2 fixed points $A(a, 0)$ and $B(-a, 0)$ is equal to a constant $2 c^2$, then the equation of its locus is

• Correct answer
  A

  $x^2+y^2=c^2-a^2$
• B

  $x^2+y^2=c^2+a^2$
• C

  $2 x^2+2 y^2=c^2+a^2$
• D

  $2 x^2-2 y^2=c^2+a^2$

Solution

Let the point be $P(x, y)$.
Then, $$ \begin{aligned} & B P=\sqrt{(x+a)^2+(y-0)^2} \\ & \text { and } A P=\sqrt{(x-a)^2+(y-0)^2} \end{aligned} $$ Given that, $$ (B P)^2+(A P)^2=2 c^2 $$ i.e. $(x+a)^2+y^2+(x-a)^2+y^2=2 c^2$ $$ \begin{aligned} & \Rightarrow x^2+a^2+2 a x+2 y^2+x^2+a^2-2 a x=2 c^2 \\ & \Rightarrow 2 x^2+2 y^2+2 a^2=2 c^2 \\ & \quad \text { or } x^2+y^2=c^2-a^2 \end{aligned} $$ Desired locus is $x^2+y^2=c^2-a^2$

Hence, the correct answer is option: 'A'.