The equation of a circle with centre C is $x^2 + y^2 - 8x - 10y + 16 = 0$. A point $B(7, 9)$ lies on the circumference of the circle. $P$ is the lowest point on the circle.

1. Group the $x$ and $y$ terms and complete the square for both

$x^2 - 8x + y^2 - 10y + 16 = 0$
$(x - 4)^2 - 4^2 + (y - 5)^2 - 5^2 + 16 = 0$

2. Rearrange into the standard circle equation form $(x-a)^2 + (y-b)^2 = r^2$

$(x - 4)^2 - 16 + (y - 5)^2 - 25 + 16 = 0$
$(x - 4)^2 + (y - 5)^2 = 25$

3. Identify the coordinates of the centre and the radius

Centre $C = (4, 5)$
Radius $r = \sqrt{25} = 5$

1. Find the gradient of the radius $CB$

From part (a), the centre of the circle is $C(4, 5)$.
Gradient of $CB = \frac{9 - 5}{7 - 4} = \frac{4}{3}$

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 $= -\frac{1}{\frac{4}{3}} = -\frac{3}{4}$

3. Find the equation of the tangent

Using the point $B(7, 9)$ and gradient $-\frac{3}{4}$:
$y - 9 = -\frac{3}{4}(x - 7)$
$4(y - 9) = -3(x - 7)$
$4y - 36 = -3x + 21$
$3x + 4y = 57$

1. Determine the y-coordinate of the lowest point on the circle

From part (a), the centre of the circle is $C(4, 5)$ and the radius is $r = 5$.
The lowest point $P$ on the circle has a y-coordinate given by $y_C - r$.
$y$-coordinate of $P = 5 - 5 = 0$