The equation of a circle with centre C is $x^{2}+y^{2}-8x-2y-152=0$.

1. Group the $x$ and $y$ terms and move the constant to the right side

$x^{2} - 8x + y^{2} - 2y = 152$

2. Complete the square for $x$ and $y$

$(x - 4)^{2} - 16 + (y - 1)^{2} - 1 = 152$
$(x - 4)^{2} + (y - 1)^{2} = 152 + 16 + 1$

3. Simplify to standard circle equation form

$(x - 4)^{2} + (y - 1)^{2} = 169$

4. Extract the centre coordinates and radius

Centre C is $(4, 1)$.
Radius $r = \sqrt{169} = 13$.

1. Substitute the coordinates of P into the left-hand side of the circle's equation.

Substitute $x = 9$ and $y = 13$ into $x^{2}+y^{2}-8x-2y-152$:
$\text{LHS} = (9)^{2} + (13)^{2} - 8(9) - 2(13) - 152$

2. Evaluate the expression to show it equals 0.

$\text{LHS} = 81 + 169 - 72 - 26 - 152$
$\text{LHS} = 250 - 250 = 0$
Since $\text{LHS} = \text{RHS}$, the point $P(9, 13)$ lies on the circle.

1. Find the coordinates of the centre C from part (i)

From the circle equation $x^{2}+y^{2}-8x-2y-152=0$, completing the square gives:
$(x-4)^2 - 16 + (y-1)^2 - 1 - 152 = 0$
$(x-4)^2 + (y-1)^2 = 169$
Centre $C$ is $(4, 1)$.

2. Find the gradient of the line l

Gradient of $CP = \frac{13 - 1}{9 - 4} = \frac{12}{5}$
The tangent at $P$ is perpendicular to the radius $CP$, so its gradient is $-\frac{5}{12}$.
Since line $l$ is parallel to the tangent, the gradient of line $l$ is also $-\frac{5}{12}$.

3. Find the coordinates of Q

Let $Q$ be $(x, y)$. Since $P(9, 13)$ is the midpoint of $QC$:
$\frac{x + 4}{2} = 9 \implies x = 14$
$\frac{y + 1}{2} = 13 \implies y = 25$
So, $Q$ is $(14, 25)$.

4. Find the equation of line l

Using the point $Q(14, 25)$ and gradient $m = -\frac{5}{12}$:
$y - 25 = -\frac{5}{12}(x - 14)$
$12y - 300 = -5x + 70$
$5x + 12y = 370$

1. Determine the perpendicular distance from the centre C to the line $l$

Since line $l$ is parallel to the tangent at P, it is perpendicular to the radius CP.
Because P is the midpoint of QC, the points C, P, and Q are collinear.
Thus, the line segment CQ is perpendicular to $l$ at Q, meaning the perpendicular distance from the centre C to line $l$ is the length CQ.

2. Compare the distance CQ with the radius of the circle to conclude

Since P is the midpoint of QC, $CQ = 2 \times CP$.
Since CP is the radius of the circle, $CQ$ is twice the radius ($CQ = 26 > 13$).
Because the perpendicular distance from the centre to line $l$ is strictly greater than the radius, the line $l$ lies entirely outside the circle and will not intersect it.