[tex]5x^2+5y^2-10x+20y-30=0[/tex]
We need to get the equation in square form
We apply completing the square method
[tex]5x^2-10x+5y^2+20y-30=0[/tex]
Divide all the terms by 5
[tex]x^2-2x+y^2+4y-6=0[/tex]
Group x terms and y terms separately
[tex](x^2-2x)+(y^2+4y)-6=0[/tex]
By completing the square method, we take half of coefficient of x and square it
[tex]\frac{-2}{2} = -1[/tex]
square it [tex](-1)^2= 1[/tex]
Add 1 on both sides
[tex](x^2-2x+1)+(y^2+4y)-6=+1[/tex]
[tex]\frac{4}{2} = 2[/tex]
square it [tex](2)^2= 4[/tex]
Add 4 on both sides
[tex](x^2-2x+1)+(y^2+4y+4)-6=0+1+4[/tex]
Add 6 on both sides
[tex](x^2-2x+1)+(y^2+4y+4)=6+1+4[/tex]
Now factor both parenthesis
[tex](x-1)^2+(y+2)^2=11[/tex]
The equation of the circle is [tex](x-1)^2+(y+2)^2=11[/tex]