Step 1: Understanding the Concept:
This question asks for the length of a tangent drawn from an external point to a circle. There is a direct formula for this calculation in coordinate geometry.
Step 2: Key Formula or Approach:
The length of the tangent, L, from an external point $(x_1, y_1)$ to a circle with the equation $S \equiv x^2 + y^2 + 2gx + 2fy + c = 0$ is given by the formula:
\[ L = \sqrt{S_1} = \sqrt{x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c} \]
This means we simply substitute the coordinates of the point into the circle's expression and take the square root.
Step 3: Detailed Explanation:
The given external point is $(x_1, y_1) = (5, 1)$.
The equation of the circle is $x^2 + y^2 + 6x - 4y - 3 = 0$.
We need to calculate $S_1$ by substituting $x=5$ and $y=1$ into the left-hand side of the circle's equation:
\[ S_1 = (5)^2 + (1)^2 + 6(5) - 4(1) - 3 \]
\[ S_1 = 25 + 1 + 30 - 4 - 3 \]
\[ S_1 = 56 - 7 \]
\[ S_1 = 49 \]
The length of the tangent is $L = \sqrt{S_1}$.
\[ L = \sqrt{49} = 7 \]
Step 4: Final Answer:
The length of the tangent is 7 units. Therefore, option (B) is correct.