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 gradient of the normal (radius)

The radius to the point of tangency is perpendicular to the tangent.
Gradient of normal $m_n = -\frac{1}{m_t} = \frac{3}{4}$

3. Find the equation of the normal

The normal passes through $A(6, 10)$.
$y - 10 = \frac{3}{4}(x - 6)$
$4y - 40 = 3x - 18$
$4y = 3x + 22$

4. Find the coordinates of the centre of the circle

The centre lies on the normal $4y = 3x + 22$ and the given line $3y = 4x + 13$.
From the normal: $x = \frac{4y - 22}{3}$
Substitute into the given line:
$3y = 4\left(\frac{4y - 22}{3}\right) + 13$
$9y = 16y - 88 + 39$
$7y = 49 \implies y = 7$
$x = \frac{4(7) - 22}{3} = \frac{6}{3} = 2$
Centre is $(2, 7)$.

5. Find the radius squared of the circle

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

6. Write the equation of the circle

Equation of the circle is $(x - a)^2 + (y - b)^2 = r^2$
$(x - 2)^2 + (y - 7)^2 = 25$