Find the equation of a circle passing through the point $(3, 4)$ and having its center at the point $(1, 2)$.

The general equation of a circle centered at $(h, k)$ with radius $r$ is:
$ (x – h)^2 + (y – k)^2 = r^2 $

Here, the center $(h, k)$ is $(1, 2)$. So the equation becomes:
$ (x – 1)^2 + (y – 2)^2 = r^2 $

Since the point $(3, 4)$ lies on the circle:
$ (3 – 1)^2 + (4 – 2)^2 = r^2 $

Simplifying, we get:
$ 2^2 + 2^2 = r^2 $
$ 4 + 4 = r^2 $
$ 8 = r^2 $

Therefore, the equation of the circle is:
$ (x – 1)^2 + (y – 2)^2 = 8 $

The equation of a circle with center $(h, k)$ and radius $r$ is:
$ (x – h)^2 + (y – k)^2 = r^2 $

Given the center $(h, k) = (1, 2)$, the equation is:
$ (x – 1)^2 + (y – 2)^2 = r^2 $

Point $(3, 4)$ lies on the circle, so:
$ (3 – 1)^2 + (4 – 2)^2 = r^2 $

Solving we get:
$ 4 + 4 = r^2 $
$ 8 = r^2 $

Hence, the circle’s equation is:
$ (x – 1)^2 + (y – 2)^2 = 8 $

For a circle with center $(h, k)$ and radius $r$, the equation is:
$ (x – h)^2 + (y – k)^2 = r^2 $

Here, $(h, k) = (1, 2)$, so:
$ (x – 1)^2 + (y – 2)^2 = r^2 $

Given point $(3, 4)$:
$ (3 – 1)^2 + (4 – 2)^2 = r^2 $

Simplified:
$ 4 + 4 = r^2 $
$ r^2 = 8 $

Final equation:
$ (x – 1)^2 + (y – 2)^2 = 8 $

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