Question

On a coordinate plane, a circle has a center at (0, 0). Point (3, 0) lies on the circle.
Distance formula: StartRoot (x 2 minus x 1) squared + (y 2 minus y 1) squared EndRoot

Does the point (2, StartRoot 6 EndRoot) lie on the circle shown? Explain.

Yes, the distance from (3, 0) to (0, 0) is 3 units.
Yes, the distance from (0, 0) to (2, StartRoot 6 EndRoot) is 3 units.
No, the distance from (3, 0) to (2, StartRoot 6 EndRoot) is not 3 units.
No, the distance from (0, 0) to (2, StartRoot 6 EndRoot) is not 3 units

Answer 1

To determine if the point \((2, \sqrt{6})\) lies on the circle with center at \((0, 0)\) and radius 3, we need to check if the distance from the point \((2, \sqrt{6})\) to the center \((0, 0)\) is equal to the radius of the circle, which is 3 units.

Using the distance formula:

\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]

we can substitute in the coordinates:

- \( (x_1, y_1) = (0, 0) \)
- \( (x_2, y_2) = (2, \sqrt{6}) \)

Now, we calculate the distance:

\[
d = \sqrt{(2 - 0)^2 + (\sqrt{6} - 0)^2}
\]
\[
= \sqrt{2^2 + (\sqrt{6})^2}
\]
\[
= \sqrt{4 + 6}
\]
\[
= \sqrt{10}
\]

The distance from the center \((0, 0)\) to the point \((2, \sqrt{6})\) is \(\sqrt{10}\), which is not equal to 3 (the radius of the circle). Thus, the point \((2, \sqrt{6})\) does not lie on the circle.

So the correct choice is:

**No, the distance from (0, 0) to (2, \sqrt{6}) is not 3 units.**