Answer:
[tex]x^{2}+y^{2}+2x+10y-5=0[/tex]
[tex]3x^{2}+3y^{2}+12x+30y-6=0[/tex]
Step-by-step explanation:
we know that
The equation of a circle ion standard form is equal to
[tex](x-h)^{2} +(y-k)^{2}=r^{2}[/tex]
where
(h,k) is the center and r is the radius
Verify each case
case 1) we have
[tex](x+5)^{2} +(y+2)^{2}=-4[/tex]
The radius squared cannot be a negative number
[tex]r^{2}\neq -4[/tex]
therefore
This equation is not the equation of a circle
case 2) we have
[tex]2x^{2}+y^{2}+15x-y+3=0[/tex]
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex](2x^{2}+15x)+(y^{2}-y)=-3[/tex]
Factor the leading coefficient of each expression
[tex]2(x^{2}+7.5x)+(y^{2}-y)=-3[/tex]
Complete the square twice. Remember to balance the equation by adding the same constants to each side.
[tex]2(x^{2}+7.5x+14.0625)+(y^{2}-y+0.25)=-3+28.125+0.25[/tex]
[tex]2(x^{2}+7.5x+14.0625)+(y^{2}-y+0.25)=25.375[/tex]
Rewrite as perfect squares
[tex]2(x+3.75)^{2}+(y-0.5)^{2}=25.375[/tex]
This equation is not the equation of a circle
case 3) we have
[tex]x^{2}+y^{2}+2x+10y-5=0[/tex]
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex](x^{2}+2x)+(y^{2}+10y)=5[/tex]
Complete the square twice. Remember to balance the equation by adding the same constants to each side.
[tex](x^{2}+2x+1)+(y^{2}+10y+25)=5+1+25[/tex]
[tex](x^{2}+2x+1)+(y^{2}+10y+25)=31[/tex]
Rewrite as perfect squares
[tex](x+1)^{2}+(y+5)^{2}=31[/tex]
This equation represent the equation of a circle
case 4) we have
[tex]3x^{2}+3y^{2}+12x+30y-6=0[/tex]
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex](3x^{2}+12x)+(3y^{2}+30y)=6[/tex]
Factor the leading coefficient of each expression
[tex]3(x^{2}+4x)+3(y^{2}+10y)=6[/tex]
Complete the square twice. Remember to balance the equation by adding the same constants to each side.
[tex]3(x^{2}+4x+4)+3(y^{2}+10y+25)=6+12+75[/tex]
[tex]3(x^{2}+4x+4)+3(y^{2}+10y+25)=93[/tex]
Rewrite as perfect squares
[tex]3(x+2)^{2}+3(y+5)^{2}=93[/tex]
Divide by 3 both sides
[tex](x+2)^{2}+(y+5)^{2}=31[/tex]
This equation represent the equation of a circle
case 5) we have
[tex]x^{2}-y^{2}+20x+2y+9=0[/tex]
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex](x^{2}+20x)-(y^{2}-2y)=-9[/tex]
Complete the square twice. Remember to balance the equation by adding the same constants to each side.
[tex](x^{2}+20x+100)-(y^{2}-2y+1)=-9+10-1[/tex]
[tex](x^{2}+20x+100)-(y^{2}-2y+1)=0[/tex]
Rewrite as perfect squares
[tex](x+10)^{2}-(y-1)^{2}=0[/tex]
This equation is not the equation of a circle
