In the diagram, M is the centre of the circle having equation (x−3)²+y²=25. E(-1; 3) and T are points on the circle. EC is a tangent to the circle at E and cuts the x-axis at D. ED= 15/4 units. MT is produced to meet the tangent at C(-7; p).

The line segment EC is a tangent to the circle at point E, and EM is the radius to the point of tangency. A tangent to a circle is always perpendicular to the radius at the point of tangency.

Therefore, the angle CÊM is $90^\circ$.

From the circle equation $(x-3)^2 + y^2 = 25$, the center is $M(3, 0)$.
The point E is given as $E(-1, 3)$.

The gradient $m_{EM} = \frac{y_2 - y_1}{x_2 - x_1}$.

Since the tangent EC is perpendicular to the radius EM, the product of their gradients is -1.

Using the point-slope form $y - y_1 = m(x - x_1)$ with $E(-1, 3)$ and $m_{EC} = \frac{4}{3}$:

D is the x-intercept of the tangent EC. Set $y=0$ in the equation of EC:

So, $D\left(-\frac{13}{4}, 0\right)$.

The center of the circle is $M(3, 0)$.

Using the distance formula $DM = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:

Point C is given as $C(-7, p)$. The equation of the tangent EC is $y = \frac{4}{3}x + \frac{13}{3}$ (from 4.2).

In a parallelogram, the diagonals bisect each other. For parallelogram SEMC, the diagonals are SM and EC. Their midpoints must be the same.

$E(-1, 3)$, $M(3, 0)$, $C(-7, -5)$ (using $p=-5$). Let $S(x_S, y_S)$.

The midpoint of SM is $\left(\frac{x_S + 3}{2}, \frac{y_S + 0}{2}\right)$.
Equate the coordinates of the midpoints:

So, $S(-11, -2)$. This satisfies the condition $x_S < 0$.

The original radius is $r = \sqrt{25} = 5$ units.
The new radius is $r_{new} = 5 + 7 = 12$ units.
The center remains $M(3, 0)$.

From 4.5, $S(-11, -2)$.

Using the distance formula $MS = \sqrt{(x_S - x_M)^2 + (y_S - y_M)^2}$:

$MS = 10\sqrt{2} \approx 14.142$ units.
The new radius $r_{new} = 12$ units.
Since $MS \approx 14.142 > 12$, point S lies outside the new circle.

M is the center, and E and T are points on the circle. Therefore, EM and MT are radii, so $EM = MT = 5$.
This means $\triangle EMT$ is an isosceles triangle.

The problem states "MT is produced to meet the tangent at C". This means M, T, and C are collinear.
Therefore, the angle $\angle ETM$ (EÎM) is the same as $\angle EMC$.

$ME = 5$ (radius).
$CE = \sqrt{(-7 - (-1))^2 + (-5 - 3)^2} = \sqrt{(-6)^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$.
$MC = \sqrt{(-7 - 3)^2 + (-5 - 0)^2} = \sqrt{(-10)^2 + (-5)^2} = \sqrt{100 + 25} = \sqrt{125} = 5\sqrt{5}$.

From 4.1, we know $\angle CEM = 90^\circ$. We can use this right triangle to find $\angle EMC$.

Since $\angle ETM = \angle EMC$,