Question

give the equation of the circle centered at the origin and passing through the point (0,4).

Explanation:

Step1: Recall the standard equation of a circle

The standard equation of a circle is \((x - h)^2+(y - k)^2=r^2\), where \((h,k)\) is the center and \(r\) is the radius.

Step2: Determine the center and radius

Since the circle is centered at the origin, \((h,k)=(0,0)\). The circle passes through \((0,4)\), so the radius \(r\) is the distance from \((0,0)\) to \((0,4)\). Using the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\), we have \(r=\sqrt{(0 - 0)^2+(4 - 0)^2}=4\).

Step3: Substitute into the standard equation

Substituting \(h = 0\), \(k = 0\), and \(r = 4\) into \((x - h)^2+(y - k)^2=r^2\), we get \((x - 0)^2+(y - 0)^2=4^2\), which simplifies to \(x^2+y^2 = 16\).

Answer:

\(x^2+y^2=16\)