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A point $P(x, y)$ is such that the sum of squares of its distance from $(a, 0)$ and $(-a, 0)$ is $2 b^2$. The equation representing the locus of $P$ is
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Verified Answer
The correct answer is:
$x^2+y^2=b^2-a^2$
Let the point be $(x, y)$.
$$
\begin{aligned}
& \Rightarrow \quad(x-a)^2+(y-0)^2+(x+a)^2+(y-0)^2=2 b^2 \\
& \Rightarrow \quad 2 x^2+2 y^2+2 a^2=2 b^2 \\
& \Rightarrow \quad x^2+y^2=b^2-a^2 .
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow \quad(x-a)^2+(y-0)^2+(x+a)^2+(y-0)^2=2 b^2 \\
& \Rightarrow \quad 2 x^2+2 y^2+2 a^2=2 b^2 \\
& \Rightarrow \quad x^2+y^2=b^2-a^2 .
\end{aligned}
$$
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