Given two pairs of co-ordinates through which the circle passes and the equation of the line through which the circle passes, we can substitute those values in the equation of the circle and the line and solve for h, k and r.

Given that the circle passes through $(4,1) \rightarrow (4-h)^2 + (1-k)^2 = r^2... (1)$

Given that the circle passes through $(6,5) \rightarrow (6-h)^2 + (5-k)^2 = r^2 ... (2)$

From $(1)$ and $(2)$, we get, $ (4-h)^2 + (1-k)^2 = (6-h)^2 + (5-k)^2 \rightarrow 4h + 8k = 44 \rightarrow h+2k = 11 ... (3)$

Since the centre of the circle lines on $4x+y=16 \rightarrow 4h+k = 16 ...(4)$

Solving $(3)$ and $(4)$, we get, $h = 3$ and $k = 4$.

Substituting the values of $h$ and $k$ in $(1)$, we get: $(4-3)^2 + (1-4)^2 = r^2 \rightarrow r^2 = 10 \rightarrow r = \sqrt{10}$

Now, we can substitute for $h, k$ and $r$ in $(x-h)^2+(y-k)^2=r^2$ to get the equation of the circle:

$\quad \quad (x-3)^2 + (y-4)^2 = \sqrt{10}^2$

$\quad \quad x^2 + y^2 -6x-8y + 15 = 0$