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: Recall the distance formula

The distance formula between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). For a point \((x,y)\) to lie on a circle centered at \((0,0)\) with radius \(r\), the distance from \((0,0)\) to \((x,y)\) must be equal to \(r\). From the graph, the radius \(r = 3\) (distance from \((0,0)\) to \((3,0)\)).

Step2: Calculate the distance from \((0,0)\) to \((2,\sqrt{6})\)

Substitute \(x_1 = 0,y_1 = 0,x_2=2,y_2=\sqrt{6}\) into the distance formula:

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Wait, no! Wait, let's re - check. Wait, no, wrong. Wait, the radius is 3. The standard equation of a circle centered at \((h,k)=(0,0)\) is \(x^{2}+y^{2}=r^{2}\). Here \(r = 3\), so the equation is \(x^{2}+y^{2}=9\). For the point \((x = 2,y=\sqrt{6})\), substitute into the left - hand side of the equation: \(x^{2}+y^{2}=2^{2}+(\sqrt{6})^{2}=4 + 6=9\).

Alternatively, using the distance formula correctly:
Substitute \(x_1=0,y_1 = 0,x_2 = 2,y_2=\sqrt{6}\) into \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\)

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Answer:

Yes, the distance from \((0,0)\) to \((2,\sqrt{6})\) is 3 units.