Determine whether the point $(6, 0)$ is on, inside, or outside the circle defined by the equation $(x - 5)^2 + (y - 1)^2 = 9$.

1. Problem Description

Determine whether the point $(6, 0)$ is on, inside, or outside the circle defined by the equation $(x - 5)^2 + (y - 1)^2 = 9$.

2. Solution Steps

The equation of a circle with center $(h, k)$ and radius $r$ is given by:

$(x - h)^2 + (y - k)^2 = r^2$.

In this case, the equation is $(x - 5)^2 + (y - 1)^2 = 9$.

So, the center of the circle is $(5, 1)$ and the radius is $r = \sqrt{9} = 3$.

To determine the position of the point $(6, 0)$ with respect to the circle, we can calculate the distance between the point and the center of the circle and compare it to the radius.

The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

In our case, $(x_1, y_1) = (5, 1)$ (the center of the circle) and $(x_2, y_2) = (6, 0)$.

$d = \sqrt{(6 - 5)^2 + (0 - 1)^2} = \sqrt{(1)^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2}$.

Since $r = 3$ and $d = \sqrt{2}$, we compare $d$ and $r$.

Since $\sqrt{2} \approx 1.414$ and $3 > \sqrt{2}$, we have $r > d$.

Therefore, the point $(6, 0)$ is inside the circle.

3. Final Answer

Inside the circle.