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, \( x=0, y=\pm \sqrt{7} \).