The centre of a circle is $(-6,4)$. If one end of the diameter of the circle is at $(-12,8)$, then the other end is at

The centre is the midpont of the line joining the two points of the diameter.

Using Midpoint-formula $\left(\alpha,\beta\right)\equiv\left(\dfrac{{x}_{1}+{x}_{2}}{2},\dfrac{{y}_{1}+{y}_{2}}{2}\right)$

From above we have $\alpha=-6,\beta=4$

Let ${x}_{1}=x,{y}_{1}=y$

Given${x}_{2}=-12,{y}_{2}=8$

We have $\alpha=\dfrac{{x}_{1}+{x}_{2}}{2}$ and $\beta=\dfrac{{y}_{1}+{y}_{2}}{2}$

$x+\dfrac{\left(-12\right)}{2} = -6$

$\Rightarrow x-6=-6$

$\therefore x = 0$

$\Rightarrow y+\dfrac{8}{2} = 4$

$\Rightarrow y+4=4$

$\therefore y = 0$

Hence, the other point is at $\left(0,0\right)$