Answer:
• Centre of the circle, (h,k)=(-1,-3)
,• Radius = 6
Explanation:
The standard form of the equation of a circle is:
[tex]\begin{gathered} (x-h)^2+(y-k)^2=r^2 \\ \text{Centre}=(h,k) \end{gathered}[/tex]Given the circle below:
[tex]x^2+y^2+2x+6y=26[/tex]First, we reorder the terms.
[tex]x^2+2x+y^2+6y=26[/tex]Next, we complete the square for the quadratics in x and y as shown below:
[tex]\begin{gathered} x^2+2x+1^2+y^2+6y+3^2=26+1^2+3^2 \\ (x+1)^2+(y+3)^2=36 \\ (x+1)^2+(y+3)^2=6^2 \end{gathered}[/tex]Comparing with the standard form given above:
[tex]\begin{gathered} h=-1 \\ k=-3 \\ \text{Centre of the circle, (h,k)=(-1,-3)} \\ \text{Radius, r=6} \end{gathered}[/tex]