Given: The equation of the circle is

 $x^2+y^2-2x-6y-15=0$

$\Rightarrow {\left(x-1\right)}^2+{\left(y-3\right)}^2-25=0$

So the centre of the circle is $\left(1,3\right)$ and its radius is $5$.

We know that the centre is mid-point of the diameter.

Also one end of the diameter is $\left(4,1\right)$ and let other end be $\left(a,b\right)$.

By mid point formula, 

$1=\frac{a+4}{2}$ and $3=\frac{b+1}{2}$

$\Rightarrow a=-2$ and $b=5$

So the other end of the diameter is $\left(-2, 5\right)$.