A(1, 7) and B(9, 7) lie on a circle $C_1$ with centre P. The line $4y + 3x = 31$ is a normal to the circle and passes through A. A larger circle $C_2$ has the same centre $P$ and radius three times that of $C_1$. $T$ is a point on $C_2$ such that $P$, $A$ and $T$ are collinear and the x-coordinate of $T$ is negative.

1. Find the midpoint of the chord AB

The points $A(1, 7)$ and $B(9, 7)$ have the same y-coordinate, so $AB$ is a horizontal chord.
Midpoint of $AB = \left(\frac{1+9}{2}, \frac{7+7}{2}\right) = (5, 7)$.

2. Use the perpendicular bisector property to explain the x-coordinate of P

A geometrical property of circles states that the perpendicular bisector of any chord passes through the centre of the circle.
The perpendicular bisector of the horizontal chord $AB$ is the vertical line passing through its midpoint, which is $x = 5$.
Since the centre $P$ lies on this line, the x-coordinate of $P$ must be $5$.

1. Find the y-coordinate of the centre P

Since the normal to the circle passes through the centre, P lies on the line $4y + 3x = 31$.
Substitute $x = 5$ (from part i) into the equation:
$4y + 3(5) = 31$
$4y + 15 = 31$
$4y = 16$
$y = 4$
The centre P is $(5, 4)$.

2. Calculate the square of the radius of the circle

The radius $r$ is the distance from the centre P$(5, 4)$ to the point A$(1, 7)$ on the circle.
$r^2 = (5 - 1)^2 + (4 - 7)^2$
$r^2 = 4^2 + (-3)^2$
$r^2 = 16 + 9 = 25$

3. State the equation of the circle

The equation of the circle is $(x - a)^2 + (y - b)^2 = r^2$.
$(x - 5)^2 + (y - 4)^2 = 25$

1. Find the gradient of the normal at $A$.

The equation of the normal is $4y + 3x = 31$.
Rearranging to $y = mx + c$ form:
$4y = -3x + 31$
$y = -\frac{3}{4}x + \frac{31}{4}$
The gradient of the normal is $-\frac{3}{4}$.

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

Since the tangent is perpendicular to the normal at the point of contact $A$, the gradient of the tangent $m_t$ satisfies:
$m_t \times \left(-\frac{3}{4}\right) = -1$
$m_t = \frac{4}{3}$

3. Find the equation of the tangent and the coordinates of $D$.

The tangent passes through $A(1, 7)$. Its equation is:
$y - 7 = \frac{4}{3}(x - 1)$
$y = \frac{4}{3}x - \frac{4}{3} + 7$
$y = \frac{4}{3}x + \frac{17}{3}$

The tangent intersects the y-axis at $D$, so $x = 0$.
$y = \frac{17}{3}$
Therefore, the coordinates of $D$ are $\left(0, \frac{17}{3}\right)$.

1. Find the coordinates of the centre $P$.

From (i), the $x$-coordinate of $P$ is $5$.
Since the normal $4y + 3x = 31$ passes through the centre $P$, substitute $x = 5$:
$4y + 3(5) = 31$
$4y = 16 \implies y = 4$
So, $P$ is $(5, 4)$.

2. Determine the relationship between the positions of $P$, $A$, and $T$.

The radius of $C_1$ is $PA = \sqrt{(1 - 5)^2 + (7 - 4)^2} = \sqrt{16 + 9} = 5$.
The radius of $C_2$ is $PT = 3 \times 5 = 15$.
Since $P$, $A$, and $T$ are collinear, $T$ lies on the line $PA$. The distance $PT$ is $3$ times the distance $PA$.
Considering the horizontal and vertical changes from $P$ to $A$:
$\Delta x = 1 - 5 = -4$
$\Delta y = 7 - 4 = 3$

3. Calculate the coordinates of $T$.

To find $T$, we can multiply the changes by $\pm 3$ since $PT = 3PA$.
If $T$ is in the same direction as $A$ from $P$:
$x_T = 5 + 3(-4) = -7$
$y_T = 4 + 3(3) = 13$
This gives $T(-7, 13)$, which has a negative $x$-coordinate as required.
(If $T$ were in the opposite direction, $x_T = 5 - 3(-4) = 17$, which is rejected).