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Let the radius of the circle be $r$.
Since the center of the circle is at $(1, 2)$, the equation of the circle is:
$(x - 1)^2 + (y - 2)^2 = r^2$
$x^2 + y^2 - 2x - 4y + 1 + 4 - r^2 = 0$
$x^2 + y^2 - 2x - 4y + (5 - r^2) = 0$

The length of the tangent drawn from the origin $(0, 0)$ to this circle is given by:
$\text{Length of tangent} = \sqrt{0^2 + 0^2 - 2(0) - 4(0) + (5 - r^2)} = \sqrt{5 - r^2}$

According to the question:
$\sqrt{5 - r^2} = 2$
$5 - r^2 = 4$
$r^2 = 1$

Substituting $r^2 = 1$ into the equation of the circle:
$(x - 1)^2 + (y - 2)^2 = 1$
$x^2 - 2x + 1 + y^2 - 4y + 4 = 1$
$x^2 + y^2 - 2x - 4y + 4 = 0$

Thus, the equation of the circle is $x^2 + y^2 - 2x - 4y + 4 = 0$ (or $(x-1)^2 + (y-2)^2 = 1$).