Find the center and radius of the circle given by the equation x2 − 6x + y2 − 10y + 30 = 0.

The given equation of the circle is $x^2 - 6x + y^2 - 10y + 30 = 0$.
Group the x-terms and y-terms, and move the constant term to the right side of the equation:

$$(x^2 - 6x) + (y^2 - 10y) = -30$$

To complete the square for $x^2 - 6x$, take half of the coefficient of $x$ (which is $-6$), square it, and add it to both sides of the equation.
Half of $-6$ is $-3$. $(-3)^2 = 9$.

$$(x^2 - 6x + 9) + (y^2 - 10y) = -30 + 9$$

To complete the square for $y^2 - 10y$, take half of the coefficient of $y$ (which is $-10$), square it, and add it to both sides of the equation.
Half of $-10$ is $-5$. $(-5)^2 = 25$.

$$(x^2 - 6x + 9) + (y^2 - 10y + 25) = -30 + 9 + 25$$

The expressions in parentheses are now perfect squares:

$$(x - 3)^2 + (y - 5)^2 = 4$$

The standard form of the equation of a circle is $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ is the center and $r$ is the radius.
Comparing our equation $(x - 3)^2 + (y - 5)^2 = 4$ with the standard form:
The center $(h, k)$ is $(3, 5)$.
The radius squared $r^2 = 4$, so the radius $r = \sqrt{4} = 2$.

The center of the circle is $\boxed{(3, 5)}$.
The radius of the circle is $\boxed{2}$.