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Show that if $x^2+y^2=1$, then the point $\left(x, y, \sqrt{1-x^2-y^2}\right)$ is at a distance 1 unit from the origin.
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Verified Answer
$\quad$ Required distance $=\mathrm{d}$
$$
=\sqrt{(x-0)^2+(y-0)^2+\left(\sqrt{1-\left(x^2+y^2\right.}-0\right)^2}
$$
Now, $x^2+y^2=1$ (given)
$$
\begin{aligned}
&\therefore \mathrm{d}=\sqrt{\mathrm{x}^2+\mathrm{y}^2+(\sqrt{1-1})^2} \\
&\Rightarrow \mathrm{d}=\sqrt{1+0}=\sqrt{1}=1
\end{aligned}
$$
$$
=\sqrt{(x-0)^2+(y-0)^2+\left(\sqrt{1-\left(x^2+y^2\right.}-0\right)^2}
$$
Now, $x^2+y^2=1$ (given)
$$
\begin{aligned}
&\therefore \mathrm{d}=\sqrt{\mathrm{x}^2+\mathrm{y}^2+(\sqrt{1-1})^2} \\
&\Rightarrow \mathrm{d}=\sqrt{1+0}=\sqrt{1}=1
\end{aligned}
$$
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