1 Give a reason why TS ewline extyen NP. 4.2 Determine the equation of the line passing through N and P in the form y=mx+c.

The problem states that $T(0; 5)$ is the center of the circle. The line through $P$ and $S(-3; 8)$ intersects the circle at $N$. It is also given that $NS = PS$.

$NP$ is a chord of the circle. $S$ is the midpoint of the chord $NP$ because $NS = PS$. The line segment $TS$ connects the center of the circle ($T$) to the midpoint of the chord ($S$). A line from the center of a circle to the midpoint of a chord is perpendicular to the chord.

Reason: The line segment from the center of a circle to the midpoint of a chord is perpendicular to the chord.
Therefore, $\boxed{TS \perp NP}$.

Given points $T(0; 5)$ and $S(-3; 8)$.

From question 4.1, we know that $TS \perp NP$. For perpendicular lines, the product of their gradients is $-1$.

The line $NP$ passes through $S(-3; 8)$ and has a gradient $m_{NP} = 1$. Using the point-slope form $y - y_1 = m(x - x_1)$:

The equation of the line passing through $N$ and $P$ is $\boxed{y = x + 11}$.

Point $P$ is where the circle cuts the y-axis, and it lies on the line $y = x + 11$. For a point on the y-axis, $x=0$.
Substitute $x=0$ into the equation of line $NP$:

So, $P(0; 11)$.

The center of the circle is $T(0; 5)$ and $P(0; 11)$ is a point on the circle. The radius $r$ is the distance $TP$.

Tangents parallel to the x-axis are horizontal lines of the form $y = k$. These tangents occur at the highest and lowest points of the circle.
The y-coordinate of the center is $y_T = 5$. The radius is $r = 6$.
The highest point is $y_T + r = 5 + 6 = 11$.
The lowest point is $y_T - r = 5 - 6 = -1$.
The equations of the tangents are $\boxed{y = 11 \text{ and } y = -1}$.

Point $M$ is where the line $NP$ intersects the x-axis. For a point on the x-axis, $y=0$.
Substitute $y=0$ into the equation of line $NP$ ($y = x + 11$):

So, $M(-11; 0)$.

Given points $M(-11; 0)$ and $T(0; 5)$.

The length of $MT$ is $\boxed{\sqrt{146}}$.

The points are $S(-3; 8)$, $T(0; 5)$, and $M(-11; 0)$.
From question 4.1, we established that $TS \perp NP$. Since $M$ lies on the line $NP$, it means $TS \perp SM$.
Therefore, $\angle TSM = 90^\circ$.

If an angle inscribed in a circle is $90^\circ$, then the chord subtending that angle is the diameter of the circle. Since $\angle TSM = 90^\circ$, the segment $TM$ is the diameter of the circle passing through $S, T, M$.

The center of the circle is the midpoint of its diameter $TM$.
Let the center be $(a, b)$.

The center of the circle is $\left(-\frac{11}{2}; \frac{5}{2}\right)$.

The diameter is $MT$. From question 4.4, $MT = \sqrt{146}$.
The radius $r$ is half the diameter: $r = \frac{\sqrt{146}}{2}$.

The equation of a circle is $(x-a)^2 + (y-b)^2 = r^2$.
Substituting the center $\left(-\frac{11}{2}; \frac{5}{2}\right)$ and $r^2 = \frac{73}{2}$:

The equation of the circle STM is $\boxed{\left(x + \frac{11}{2}\right)^2 + \left(y - \frac{5}{2}\right)^2 = \frac{73}{2}}$.