Question

give the equation of the circle centered at the origin and passing through the point (9, 0).

Explanation:

Step1: Recall the standard equation of a circle

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

Step2: Determine the center and radius

The circle is centered at the origin, so \(h = 0\) and \(k = 0\). It passes through the point \((9, 0)\), so the radius \(r\) is the distance from the origin \((0, 0)\) to the point \((9, 0)\). Using the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\), we have \(r=\sqrt{(9 - 0)^2+(0 - 0)^2}=\sqrt{81 + 0}=9\).

Step3: Substitute the values into the equation

Substituting \(h = 0\), \(k = 0\), and \(r = 9\) into the standard equation, we get \((x - 0)^2 + (y - 0)^2 = 9^2\), which simplifies to \(x^2 + y^2 = 81\).

Answer:

\(x^2 + y^2 = 81\)