Question

does the point $(2, \sqrt{6})$ 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 $(2, \sqrt{6})$ is 3 units.
○ no, the distance from (3, 0) to $(2, \sqrt{6})$ is not 3 units.
○ no, the distance from (0, 0) to $(2, \sqrt{6})$ is not 3 units.
distance formula: $sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Explanation:

Step1: Identify circle radius

The circle has center $(0,0)$ and passes through $(3,0)$, so radius $r = 3$.

Step2: Calculate distance from center to point

Use distance formula for $(0,0)$ and $(2, \sqrt{6})$:
$\text{Distance} = \sqrt{(2 - 0)^2 + (\sqrt{6} - 0)^2}$

Step3: Simplify the expression

$\sqrt{2^2 + (\sqrt{6})^2} = \sqrt{4 + 6} = \sqrt{10} \approx 3.16$

Step4: Compare to radius

$\sqrt{10}
eq 3$, so the point is not on the circle.

Answer:

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