# Find the values of $x, y$ if the distances of the point $(x, y)$ from $(-3, 0)$ as well as from $(3, 0)$ are 4.

Given:

The distances of the point $(x, y)$ from $(-3, 0)$ as well as from $(3, 0)$ are 4.

To do:

We have to find the values of $x, y$.

Solution:

We know that,

The distance between two points $\mathrm{A}\left(x_{1}, y_{1}\right)$ and $\mathrm{B}\left(x_{2}, y_{2}\right)$ is $\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$.

Therefore,

The distance between $(x, y)$ and $(-3, 0)$ $=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

$=\sqrt{(-3-x)^{2}+(0-y)^{2}}$
$=\sqrt{(x+3)^{2}+y^{2}}$
The distance between $(x, y)$ and $(3,0)$ $=\sqrt{(3-x)^{2}+(0-y)^{2}}=\sqrt{(3-x)^{2}+y^{2}}$
According to the question,
$\sqrt{(x+3)^{2}+y^{2}}=4$ and $\sqrt{(3-x)^{2}+y^{2}}=4$
Squaring on both sides, we get,

$(x+3)^{2}+y^{2}=16$......(i) and $(3-x)^{2}+y^{2}=16$......(ii)
Subtracting (ii) from (i), we get,

$(x+3)^{2}-(3-x)^{2}+y^{2}-y^{2}=16-16$

$(x+3)^{2}-(3-x)^{2}=0$

$x^{2}+6 x+9-x^{2}+6 x-9=0$

$\Rightarrow 12 x=0$
$\therefore x=0$

Substituting the value of $x$ in $(i)$
$\sqrt{(x+3)^{2}+y^{2}}=(4)$

Squaring on both sides, we get,

$\Rightarrow(x+3)^{2}+y^{2}=16$
$\Rightarrow(0+3)^{2}+y^{2}=16$

$\Rightarrow 9 + y^{2}=16$
$\Rightarrow y^{2}=16-9=7$
$\therefore y=\pm \sqrt{7}$
Hence, $x=0, y=\pm \sqrt{7}$.

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Updated on: 10-Oct-2022

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