Question 1210473
This problem is a classic application of the **Tangent-Secant Theorem** (a form of the Power of a Point Theorem) in circle geometry. Since no figure is provided, we must assume the geometric setup that allows for a unique solution using the given segment lengths.

### 1. Assuming the Geometric Setup

We assume the following configuration:
* $\overline{PT}$ is a **tangent** to a circle at point $T$.
* $\overline{PSQ}$ is a **secant** line intersecting the same circle at points $S$ and $Q$.
* $P$ is the external point from which the tangent and secant originate.
* The length $ST=2$ is **extraneous information** (a common feature in test questions to check if the student uses the correct theorem).

### 2. Applying the Tangent-Secant Theorem

The theorem states that the square of the length of the tangent segment is equal to the product of the length of the external secant segment and the entire secant segment.

$$\mathbf{PT^2 = PS \cdot PQ}$$

From the given information:
* $PT = 5$
* $QS = 4$
* $PQ$ is the entire secant segment: $PQ = PS + QS = PS + 4$

### 3. Calculation

Let $PS = x$. The equation becomes:
$$5^2 = x \cdot (x + 4)$$
$$25 = x^2 + 4x$$

Rearrange into a quadratic equation:
$$x^2 + 4x - 25 = 0$$

Use the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, to solve for $x$:
$$x = \frac{-4 \pm \sqrt{4^2 - 4(1)(-25)}}{2(1)}$$
$$x = \frac{-4 \pm \sqrt{16 + 100}}{2}$$
$$x = \frac{-4 \pm \sqrt{116}}{2}$$

Simplify $\sqrt{116} = \sqrt{4 \cdot 29} = 2\sqrt{29}$:
$$x = \frac{-4 \pm 2\sqrt{29}}{2}$$
$$x = -2 \pm \sqrt{29}$$

Since $x$ represents the length $PS$, it must be positive.
$$PS = x = -2 + \sqrt{29}$$

### 4. Finding PQ

The length $PQ$ is the sum of the external segment $PS$ and the internal segment $QS$:
$$PQ = PS + QS$$
$$PQ = (-2 + \sqrt{29}) + 4$$
$$PQ = \mathbf{2 + \sqrt{29}}$$