Which point is on the circle centered at the origin with a radius of 5 units? Distance formula: StartRoot (x 2 minus x 1) squared + (y 2 minus y 1) squared EndRoot

Respuesta :

Answer:

Option A) [tex](2,\sqrt{21})[/tex]

Step-by-step explanation:

The following information is missing in the question:

A. [tex](2,\sqrt{21})[/tex]

B. [tex](2,\sqrt{23})[/tex]

C. (2, 1)

D. (2, 3)

We are given the following in the question:

A circle centered at origin and radius 5 units.

We have to find the equation of a point that lies on the circle.

Let (x,y) lie on the circle.

Distance formula:

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

Putting

[tex](x_2,y_2) = (x,y)\\(x_1.y_1) = (0,0)\\d = 5[/tex]

We get,

[tex]5 = \sqrt{(y-0)^2 + (x-0)^2}\\\sqrt{x^2+y^2}=5\\x^2+y^2 = 25[/tex]

is the required equation of point on the circle centered at the origin with a radius of 5 units.

The point [tex](2,\sqrt{21})[/tex] satisfies the given equation.

Verification:

[tex](2)^2 + (\sqrt{21})^2\\=4 + 21\\=25[/tex]

Thus, [tex](2,\sqrt{21})[/tex] lies on the circle centered at the origin with a radius of 5 units.

Answer:

A. (2,\sqrt{21})

Step-by-step explanation: