Respuesta :

Given

[tex]x^2+2x+y^2-6y=35[/tex]

Complete the square for x, as shown below

[tex]\begin{gathered} x^2+2x+b=(x+a)(x+a)=x^2+2ax+a^2 \\ \Rightarrow2=2a\Rightarrow a=1 \\ \Rightarrow b=a^2=1 \end{gathered}[/tex]

Therefore,

[tex]\begin{gathered} x^2+2x+1+y^2-6y=35+1=36 \\ \Rightarrow(x+1)^2+y^2-6y=36 \end{gathered}[/tex]

Completing the square for y,

[tex]\begin{gathered} y^2-6y+b=(x+a)^2=x^2+2ax+a^2 \\ \Rightarrow-6=2a\Rightarrow a=-3 \\ \Rightarrow b=a^2=9 \end{gathered}[/tex]

Then,

[tex](x+1)^2+(y-3)^2=45[/tex]

The answer is (x+1)^2+(y-3)^2=45. A circumference in its standard form.