Answer:
The centre of the circle is (-4,1).
Explanation
Given the equation of the circle:
[tex]\mleft(x+4\mright)^2+(y-1)^2=32[/tex]Comparing with the standard form of the equation of a circle:
[tex]\begin{gathered} (x-h)^2+(y-k)^2=r^2 \\ Where\; Centre=(h,k) \end{gathered}[/tex]We see that:
[tex]\begin{gathered} x-h=x+4 \\ \implies h=-4 \\ \text{Also:} \\ y-k=y-1 \\ \implies k=1 \end{gathered}[/tex]The centre of the circle is (-4,1).
