Given, center of the circle (-2,-5)
The radius is r=6
Now the form of the equation of circle is:
[tex](x-h)^2+(y-k)^2=r^2[/tex]
Thus,
[tex]\begin{gathered} (x-(-2))^2+(y-(-5))^2=6^2 \\ \Rightarrow(x+2)^2+(y+5)^2=36 \\ \Rightarrow x^2+4+4x+y^2+25+10y=36 \\ \Rightarrow x^2+y^2+4x+10y+29=36 \\ \Rightarrow x^2+y^2+4x+10y-7=0 \end{gathered}[/tex]
The answer is
[tex]x^2+y^2+4x+10y-7=0[/tex]
