Does the point (-1, V15) lie on the circle centered at the origin and containing the point (0,4)?Explain.

Explanation:
The equation for a circle centered at the origin is:
[tex]x^2+y^2=r^2[/tex]Where r is the radius of the circle.
For this problem it says that it contains point (0, 4). Therefore the radius of the circle is 4 and the equation is:
[tex]\begin{gathered} x^2+y^2=4^2 \\ x^2+y^2=16 \end{gathered}[/tex]If we replace the given point into the equation and the equality is true, then the point lies on the circle:
[tex]\begin{gathered} (-1)^2+(\sqrt[]{15})^2=16 \\ 1+15=16 \\ 16=16\text{ true} \end{gathered}[/tex]Answer:
Yes, because the equation for the circle is x² + y² = 16 and (-1)² + (√15)² = 16