Respuesta :

[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]

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