Question

does the point ((-1, sqrt{15})) lie on the circle centered at the origin and containing the point ((0, 4))? explain. no, because the equation for the circle is (x^2 + y^2 = 4) and ((-1)^2 + (sqrt{15})^2
eq 4). no, because the equation for the circle is (x^2 + y^2 = 16) and ((-1)^2 + (sqrt{15})^2
eq 16). yes, because the equation for the circle is (x^2 + y^2 = 4) and ((-1)^2 + (sqrt{15})^2 = 4). yes, because the equation for the circle is (x^2 + y^2 = 16) and ((-1)^2 + (sqrt{15})^2 = 16).

Explanation:

Step1: Find the circle's equation

The circle is centered at the origin \((0,0)\) and contains \((0,4)\). The radius \(r\) is the distance from \((0,0)\) to \((0,4)\), so \(r = 4\). The equation of a circle centered at the origin is \(x^{2}+y^{2}=r^{2}\), so here \(x^{2}+y^{2}=4^{2}=16\).

Step2: Check if the point satisfies the equation

Substitute \(x=-1\) and \(y = \sqrt{15}\) into \(x^{2}+y^{2}\):
\((-1)^{2}+(\sqrt{15})^{2}=1 + 15=16\), which equals \(r^{2}=16\). So the point lies on the circle.

Answer:

Yes, because the equation for the circle is \(x^{2}+y^{2}=16\) and \((-1)^{2}+(\sqrt{15})^{2}=16\).