The diagram above shows an arch bridge. The arc $AFTEB$ is part of a circle with centre $P$, and $AB$ is a chord 8 m. Point $T$ is 1 m vertically above point $M$ which is the midpoint of $AB$. $CF$ and $DE$ are perpendicular to chord $AB$. The arc $AFTEB$ can be modelled onto the coordinate plane below, where $C$ and $D$ are (-3, 0) and (3, 0) respectively.

1. Identify the coordinates of points on the circle and the centre

Since $AB = 8$ and $M(0,0)$ is the midpoint, $B$ is $(4, 0)$.
$T$ is 1 m vertically above $M$, so $T$ is $(0, 1)$.
Let the centre $P$ be $(0, y)$.

2. Set up an equation using the property that radii are equal

The distance from the centre $P$ to any point on the circle is the radius $r$.
$PT^2 = PB^2$
$(0 - 0)^2 + (1 - y)^2 = (4 - 0)^2 + (0 - y)^2 (1 - y)^2 = 16 + y^2$

3. Solve for $y$ to find the coordinates of $P$

$1 - 2y + y^2 = 16 + y^2$
$-2y = 15$
$y = -7.5$
Therefore, $P$ is $(0, -7.5)$.

1. Find the radius of the circle

The centre of the circle is $P(0, -7.5)$ and it passes through $T(0, 1)$.
Radius $r = 1 - (-7.5) = 8.5$

2. Form the equation of the circle

Using the standard form $(x - a)^2 + (y - b)^2 = r^2$:
$(x - 0)^2 + (y - (-7.5))^2 = 8.5^2$
$x^2 + (y + 7.5)^2 = 72.25$

1. Substitute the x-coordinate of $F$ into the equation of the circle

From (ii), the equation of the circle is $x^2 + (y + 7.5)^2 = 8.5^2$.
Since $CF$ is perpendicular to the x-axis and $C$ is $(-3, 0)$, the x-coordinate of $F$ is $-3$.
Substitute $x = -3$ into the equation:
$(-3)^2 + (y + 7.5)^2 = 72.25$
$9 + (y + 7.5)^2 = 72.25$

2. Solve for the y-coordinate of $F$ to find the height

$(y + 7.5)^2 = 63.25$
$y + 7.5 = \pm \sqrt{63.25}$
Since $F$ is above the x-axis, $y > 0$:
$y = \sqrt{63.25} - 7.5$
$y \approx 0.45298$
Height of pillar $CF = 0.453\text{ m}$ (to 3 s.f.)

1. Find the gradient of the radius $PB$

The coordinates of $P$ are $(0, -7.5)$ and $B$ is $(4, 0)$.
Gradient of $PB = \frac{0 - (-7.5)}{4 - 0} = \frac{7.5}{4} = \frac{15}{8}$

2. Find the gradient of the tangent at $B$

Since the tangent is perpendicular to the radius at the point of contact,
Gradient of tangent $= -\frac{1}{\frac{15}{8}} = -\frac{8}{15}$

3. Find the equation of the tangent line

Using the point $B(4, 0)$ and gradient $-\frac{8}{15}$:
$y - 0 = -\frac{8}{15}(x - 4)$
$15y = -8(x - 4)$
$15y = -8x + 32$
$8x + 15y = 32$