which point is on the circle centered at the origin with a radius of 5 units? distance formula: $sqrt{(x_2…

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)

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)

Answer

Explanation:

Step1: Recall the equation of a circle centered at origin

The equation of a circle centered at the origin $(0,0)$ with radius $r$ is $x^{2}+y^{2}=r^{2}$. Here $r = 5$, so the equation is $x^{2}+y^{2}=25$.

Step2: Check each point

For the point $(2,\sqrt{21})$: Substitute $x = 2$ and $y=\sqrt{21}$ into the equation $x^{2}+y^{2}$. We get $2^{2}+(\sqrt{21})^{2}=4 + 21=25$. For the point $(2,\sqrt{23})$: Substitute $x = 2$ and $y=\sqrt{23}$ into the equation $x^{2}+y^{2}$, we have $2^{2}+(\sqrt{23})^{2}=4+23 = 27\neq25$. For the point $(2,1)$: Substitute $x = 2$ and $y = 1$ into the equation $x^{2}+y^{2}$, we get $2^{2}+1^{2}=4 + 1=5\neq25$. For the point $(2,3)$: Substitute $x = 2$ and $y = 3$ into the equation $x^{2}+y^{2}$, we get $2^{2}+3^{2}=4 + 9=13\neq25$.

Answer:

$(2,\sqrt{21})$