Option D: No, the distance from [tex](0,0)[/tex] to [tex](2, \sqrt{6})[/tex] is not 3 units.
Explanation:
From the figure, we can see that the radius of the circle is 3 units.
We need to determine that the point [tex](2, \sqrt{6})[/tex] lie on the circle.
This can be determined by substituting the coordinate [tex](0,0)[/tex] and [tex](2, \sqrt{6})[/tex] in the distance formula,
[tex]d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}[/tex]
Thus, we get,
[tex]d=\sqrt{\left(\sqrt{6} -0\right)^{2}+\left(2-0\right)^{2}}[/tex]
Simplifying, we have,
[tex]d=\sqrt{\left(\sqrt{6} )^{2}+\left(2)^{2}}[/tex]
[tex]d=\sqrt{6+4}[/tex]
[tex]d=\sqrt{10}[/tex]
[tex]d=3.2[/tex]
Thus, the distance from [tex](0,0)[/tex] to [tex](2, \sqrt{6})[/tex] is 3.2 units.
Hence, the point [tex](2, \sqrt{6})[/tex] does not lie on the circle because the distance is more than the radius 3 units.
Therefore, the distance from [tex](0,0)[/tex] to [tex](2, \sqrt{6})[/tex] is not 3 units.
Thus, Option D is the correct answer.