Question

Does the point (2√3, 2) lie on the circle that is centered at the origin and contains the point (0, -4)? Explain?

Answer 1

Answer:

The answer to your question is Yes, the point P lies on the circle

Step-by-step explanation:

Data

P (2$\sqrt{3}$, 2)

Center (0, 0)

Q (0, -4)

Process

1.- Find the radius of the circle

dCQ = $\sqrt{(0-0)^{2}+ (-4 + 0)^{2}}$

dCQ = $\sqrt{0^{2} + (-4)^{2}}$

dCQ = $\sqrt{16}$

dCQ = 4

2.- Find the equation of the circle

    (x - 0)² + (y - 0)² = 4²

-Simplification

           x² + y² = 16

3.- Substitute P in the equation of the circle

          (2$\sqrt{3}$)² + (2)² = 16

           4(3) + 4 = 16

               12 + 4 = 16

                     16 = 16

4.- Conclusion

The point (2$\sqrt{3}$, 2) lies on the circle because when we evaluate the equation of the circle with this point, we get the length of the radius.