Answer:
B. Centre=(-3,-5), Radius=4
Explanation:
Given the circle described by the equation:
![x^2+y^2+6x+10y+18=0](https://img.qammunity.org/2023/formulas/mathematics/college/iorkz350dqkpbg0x6ctdz1vs91w5d8nhig.png)
In order to determine the centre and radius of the circle, we write it in the standard form below:
![(x-a)^2+(y-b)^2=r^2\text{ where }\begin{cases}center=(a,b) \\ radius=r\end{cases}](https://img.qammunity.org/2023/formulas/mathematics/college/l7490iiht8zgwdkk629gblitasyl13lp4n.png)
To do this, we use the method of completing the square for both x and y.
Begin by rearranging to bring like variables together.
![x^2+6x+y^2+10y=-18](https://img.qammunity.org/2023/formulas/mathematics/college/yb15obty2qjxuwpd7yiv0hsvl2szai96mb.png)
Next, to complete the square in each variable:
• Divide the coefficient of x by 2
,
• Square the result and add to both sides of the equation.
Do the same for y.
![\begin{gathered} x^2+6x+3^2+y^2+10y+5^2=-18+3^2+5^2 \\ \implies(x+3)^2+(y+5)^2=16 \\ \implies(x+3)^2+(y+5)^2=4^2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/bd3gmsaezx9q1jtqfs3cvwtni4z1ogj6ms.png)
Comparing with the standard form given earlier:
![\begin{gathered} \mleft(x-a\mright)^2+\mleft(y-b\mright)^2=r^2 \\ \mleft(x+3\mright)^2+\mleft(y+5\mright)^2=4^2 \\ \implies a=-3,b=-5,r=4 \\ \text{Centre}=(-3,-5),\text{Radius}=4 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/6tigj4zdpicbnvhpuc5g9xl3uhstrms6uv.png)
The centre and radius of the circle are (-3,-5) and 4 respectively.