ANSWER
![(x+3)^2+(y+4)^2=1](https://img.qammunity.org/2023/formulas/mathematics/college/ipq4lv13fl1myyjeb1fgdxaito254rak1x.png)
Step-by-step explanation
We want to write the equation of the circle in vertex form:
![x^2+y^2+6x+8y+24=0](https://img.qammunity.org/2023/formulas/mathematics/college/xnzuq5udtu2hgwhmey4dh1i7txxazpx9u8.png)
The first step is to group the x terms and y terms together and take the constant to the right-hand side of the equality sign:
![x^2+6x+y^2+8y=-24](https://img.qammunity.org/2023/formulas/mathematics/college/imksuni7y7xg7w729zlv19vfqzb5uoscp6.png)
Now, complete the square for the x terms:
![\begin{gathered} x^2+6x+((6)/(2))^2+y^2+8y=-24+((6)/(2))^2 \\ x^2+6x+9+y^2+8y=-24+9 \\ (x+3)^2+y^2+8y=-15 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/exkaa1f9rpge6t4ztenx30v56nqok9bh9s.png)
Repeat the process for the y terms:
![\begin{gathered} (x+3)^2+y^2+8y+((8)/(2))^2=-15+((8)/(2))^2 \\ (x+3)^2+y^2+8y+16=-15+16 \\ (x+3)^2+(y+4)^2=1 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/2y8uxde7qux1bpd0ny3afczy3vam0o2ya2.png)
That is the equation of the circle in vertex form.