Question

which point is on the circle centered at the origin with a radius of 5 units?
distance formula: $\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$
$(2,\sqrt{21})$
$(2,\sqrt{23})$
$(2,1)$
$(2,3)$

Explanation:

Step1: Recall the distance formula

The distance \(d\) between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). Here, \((x_1,y_1)=(0,0)\) (center of the circle) and \(d = 5\) (radius of the circle). So we need to check for each point \((x,y)\) if \(\sqrt{(x - 0)^2+(y - 0)^2}=5\), or \(x^{2}+y^{2}=25\).

Step2: Check the first point \((2,\sqrt{21})\)

Substitute \(x = 2\) and \(y=\sqrt{21}\) into \(x^{2}+y^{2}\).
\(x^{2}+y^{2}=2^{2}+(\sqrt{21})^{2}\)
\(=4 + 21\)
\(=25\)

Step3: Check the second point \((2,\sqrt{23})\)

Substitute \(x = 2\) and \(y=\sqrt{23}\) into \(x^{2}+y^{2}\).
\(x^{2}+y^{2}=2^{2}+(\sqrt{23})^{2}\)
\(=4+23\)
\(=27
eq25\)

Step4: Check the third point \((2,1)\)

Substitute \(x = 2\) and \(y = 1\) into \(x^{2}+y^{2}\).
\(x^{2}+y^{2}=2^{2}+1^{2}\)
\(=4 + 1\)
\(=5
eq25\)

Step5: Check the fourth point \((2,3)\)

Substitute \(x = 2\) and \(y = 3\) into \(x^{2}+y^{2}\).
\(x^{2}+y^{2}=2^{2}+3^{2}\)
\(=4+9\)
\(=13
eq25\)

Answer:

\((2,\sqrt{21})\)