A tangent to a circle at the point $(6, 10)$ passes through the point $(9, 6)$. The centre of the circle lies on the line $3y = 4x + 13$. Showing all your working, find the equation of the circle.

1. Find the gradient of the tangent

Let the point of tangency be $A(6, 10)$ and the other point on the tangent be $B(9, 6)$.
Gradient of tangent $m_t = \frac{6 - 10}{9 - 6} = -\frac{4}{3}$

2. Find the equation of the normal to the circle at the point of tangency

The normal is perpendicular to the tangent, so its gradient $m_n = \frac{-1}{m_t} = \frac{3}{4}$.
The normal passes through $A(6, 10)$.
Equation of normal: $y - 10 = \frac{3}{4}(x - 6)$
$4y - 40 = 3x - 18$
$4y = 3x + 22$

3. Find the coordinates of the centre of the circle

The centre of the circle lies on the normal $4y = 3x + 22$ and the given line $3y = 4x + 13$.
Solve simultaneously:
$y = \frac{3}{4}x + \frac{11}{2}$
Substitute into $3y = 4x + 13$:
$3\left(\frac{3}{4}x + \frac{11}{2}\right) = 4x + 13$
$\frac{9}{4}x + \frac{33}{2} = 4x + 13$
$\frac{33}{2} - 13 = 4x - \frac{9}{4}x$
$\frac{7}{2} = \frac{7}{4}x$
$x = 2$
Substitute $x = 2$ into $4y = 3(2) + 22$:
$4y = 28 \implies y = 7$
Centre of circle is $(2, 7)$.

4. Find the radius of the circle

The radius $r$ is the distance from the centre $(2, 7)$ to the point on the circle $(6, 10)$.
$r^2 = (6 - 2)^2 + (10 - 7)^2$
$r^2 = 4^2 + 3^2 = 16 + 9 = 25$
$r = 5$

5. Write the equation of the circle

Equation of circle with centre $(a, b)$ and radius $r$ is $(x - a)^2 + (y - b)^2 = r^2$.
$(x - 2)^2 + (y - 7)^2 = 25$