Shifting center of a circle

The standard equation of a circle is $(x - h)^2 + (y - k)^2 = r^2$.
The center is the point $(h, k)$ and $r$ is the radius, so the equation tells you exactly where the circle sits and how big it is.

Shifting the center just means changing $h$ and $k$.
A circle centered at $(1, 1)$ with radius $1.5$ is $(x - 1)^2 + (y - 1)^2 = 2.25$; moving its center to $(5, 4)$ gives $(x - 5)^2 + (y - 4)^2 = 2.25$.
The radius is untouched, so the circle is the same size, just in a new place.

Watch the signs in the equation, since the center uses the opposite sign.
In $(x - 5)^2$, the $x$-coordinate of the center is $+5$, and in $(y + 4)^2 = (y - (-4))^2$, the $y$-coordinate is $-4$.
So a $+$ inside the bracket means a negative coordinate.

To shift by a given amount, adjust each coordinate.
Moving a center right by $3$ increases $h$ by $3$; moving it down by $2$ decreases $k$ by $2$.
Rewrite $h$ and $k$ with the new values and keep $r^2$ exactly as it was.

The circle $(x - 2)^2 + (y - 3)^2 = 16$ is shifted so that its center moves by $(4, -2)$. What is the equation of the new circle?
The original center is $(2, 3)$, and shifting by $(4, -2)$ moves it to $(2 + 4, 3 - 2) = (6, 1)$.
The radius is unchanged, so the new equation is $(x - 6)^2 + (y - 1)^2 = 16$.

A circle has equation $(x + 3)^2 + (y - 5)^2 = 9$, and after a shift its equation becomes $(x - 2)^2 + (y + 1)^2 = 9$. By what vector was the center shifted?
Reading the centers from the equations, the original is $(-3, 5)$ and the new one is $(2, -1)$.
The shift is the change in each coordinate: $(2 - (-3),\ -1 - 5) = (5, -6)$.

The circle $(x - 1)^2 + (y + 2)^2 = 4$ has its center moved by $(-3, 5)$. What is the new center?
The original center is $(1, -2)$, and adding the shift gives $(1 - 3,\ -2 + 5)$.
So the new center is $(-2, 3)$.