A tangent to a circle at the point $(4, 3)$ intersects the x-axis at $x = 7$. The line with the equation $3y = 2x + 6$ is normal to the circle at another point.

1. Find the gradient of the tangent at the point $(4, 3)$

The tangent passes through $(4, 3)$ and $(7, 0)$.
Gradient of tangent, $m_T = \frac{0 - 3}{7 - 4} = \frac{-3}{3} = -1$

2. Find the gradient and equation of the normal at $(4, 3)$

Gradient of normal, $m_N = -\frac{1}{m_T} = 1$
Equation of normal at $(4, 3)$:
$y - 3 = 1(x - 4)$
$y = x - 1$

3. Find the coordinates of the centre of the circle

The centre of the circle lies on the intersection of any two normals.
Given normal: $3y = 2x + 6$
Substitute $y = x - 1$ into the given normal:
$3(x - 1) = 2x + 6$
$3x - 3 = 2x + 6$
$x = 9$
$y = 9 - 1 = 8$
Centre of the circle is $(9, 8)$.

4. Calculate the square of the radius

Radius $r$ is the distance from the centre $(9, 8)$ to the point $(4, 3)$ on the circle.
$r^2 = (9 - 4)^2 + (8 - 3)^2$
$r^2 = 5^2 + 5^2 = 25 + 25 = 50$

5. Write the equation of the circle

Equation of the circle:
$(x - 9)^2 + (y - 8)^2 = 50$

1. Find the center and radius of the circle from the information in part (a)

Gradient of tangent at $(4, 3)$ passing through $(7, 0)$ is $m_t = \frac{3-0}{4-7} = -1$.
Gradient of normal at $(4, 3)$ is $m_n = 1$.
Equation of normal: $y - 3 = 1(x - 4) \implies y = x - 1$.
The other normal is $3y = 2x + 6 \implies y = \frac{2}{3}x + 2$.
Solving simultaneously for the center:
$x - 1 = \frac{2}{3}x + 2 \implies \frac{1}{3}x = 3 \implies x = 9$.
$y = 9 - 1 = 8$.
Center is $(9, 8)$.
Radius $r = \sqrt{(9-4)^2 + (8-3)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}$.

2. Determine the equations of the horizontal tangents

Tangents parallel to the x-axis are horizontal lines of the form $y = c$.
The distance from the center $(9, 8)$ to these tangents is equal to the radius $r = 5\sqrt{2}$.
Therefore, the y-coordinates of the tangents are $y = 8 \pm 5\sqrt{2}$.