A circle has a center of (2,2) and goes through the point (6,3).  Find the equation of the circle.

To find the equation of the circle, we need to determine its radius and use the standard form of the equation of a circle, which is:

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

where $(h, k)$ is the center of the circle and $r$ is the radius.

Given:

• The center of the circle is $(2, 2)$.
• The circle passes through the point $(6, 3)$.

First, we need to find the radius $r$. The radius is the distance between the center of the circle and any point on the circle. We can use the distance formula to find this distance:

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

Substitute the given points $(2, 2)$ and $(6, 3)$ into the distance formula:

$r = \sqrt{(6 - 2)^2 + (3 - 2)^2}$

Calculate the differences:

$r = \sqrt{(4)^2 + (1)^2}$

Square the differences:

$r = \sqrt{16 + 1}$

Add the squared differences:

$r = \sqrt{17}$

So, the radius $r$ is $\sqrt{17}$.

Now, we can write the equation of the circle using the center $(2, 2)$ and the radius $\sqrt{17}$:

$(x - 2)^2 + (y - 2)^2 = (\sqrt{17})^2$

Simplify the equation:

$(x - 2)^2 + (y - 2)^2 = 17$

Therefore, the equation of the circle is:

$(x - 2)^2 + (y - 2)^2 = 17$