Question

Match the circle equations in general form with their corresponding equations in standard form.

Answer 1

Answer:

the answer case A) is

x^{2} +y^{2} -4x+12y-20=0  ----->  (x-2)^{2}+(y+6)^{2}=60

Step-by-step explanation:

Answer 2

case A) $x^{2} +y^{2} -4x+12y-20=0$

Convert to standard form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$(x^{2}-4x)+(y^{2}+12y)=20$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

$(x^{2}-4x+4)+(y^{2}+12y+36)=20+4+36$

$(x^{2}-4x+4)+(y^{2}+12y+36)=60$

Rewrite as perfect squares

$(x-2)^{2}+(y+6)^{2}=60$

therefore

the answer case A) is

$x^{2} +y^{2} -4x+12y-20=0$  ----->  $(x-2)^{2}+(y+6)^{2}=60$

case B) $x^{2} +y^{2} +6x-8y-10=0$

Convert to standard form    

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$(x^{2}+6x)+(y^{2}-8y)=10$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

$(x^{2}+6x+9)+(y^{2}-8y+16)=10+9+16$

$(x^{2}+6x+9)+(y^{2}-8y+16)=35$

Rewrite as perfect squares

$(x+3)^{2}+(y-4)^{2}=35$

therefore

the answer case B) is

$x^{2} +y^{2} +6x-8y-10=0$ -----> $(x+3)^{2}+(y-4)^{2}=35$

case C) $3x^{2} +3y^{2} +12x+18y-15=0$

Convert to standard form    

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$(3x^{2}+12x)+(3y^{2}+18y)=15$

Factor the leading coefficient of each expression

$3(x^{2}+4x)+3(y^{2}+6y)=15$

$(x^{2}+4x)+(y^{2}+6y)=5$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

$(x^{2}+4x+4)+(y^{2}+6y+9)=5+4+9$

$(x^{2}+4x+4)+(y^{2}+6y+9)=18$

Rewrite as perfect squares

$(x+2)^{2}+(y+3)^{2}=18$

therefore

the answer case C) is

$3x^{2} +3y^{2} +12x+18y-15=0$ -----> $(x+2)^{2}+(y+3)^{2}=18$

case D) $5x^{2} +5y^{2} -10x+20y-30=0$

Convert to standard form    

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$(5x^{2}-10x)+(5y^{2}+20y)=30$

Factor the leading coefficient of each expression

$5(x^{2}-2x)+5(y^{2}+4y)=30$

$(x^{2}-2x)+(y^{2}+4y)=6$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

$(x^{2}-2x+1)+(y^{2}+4y+4)=6+1+4$

$(x^{2}-2x+1)+(y^{2}+4y+4)=11$

Rewrite as perfect squares

$(x-1)^{2}+(y+2)^{2}=11$    

therefore

the answer case D) is

$3x^{2} +3y^{2} +12x+18y-15=0$ -----> $(x-1)^{2}+(y+2)^{2}=11$  

case E) $2x^{2} +2y^{2} -24x-16y-8=0$  

Convert to standard form    

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$(2x^{2}-24x)+(2y^{2}-16y)=8$

Factor the leading coefficient of each expression

$2(x^{2}-12x)+2(y^{2}-8y)=8$

$(x^{2}-12x)+(y^{2}-8y)=4$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

$(x^{2}-12x+36)+(y^{2}-8y+16)=4+36+16$

$(x^{2}-12x+36)+(y^{2}-8y+16)=56$

Rewrite as perfect squares

$(x-6)^{2}+(y-4)^{2}=56$    

therefore

the answer case E) is

$2x^{2} +2y^{2} -24x-16y-8=0$ ----->  $(x-6)^{2}+(y-4)^{2}=56$    

case F) $x^{2} +y^{2}+2x-12y-9=0$  

Convert to standard form    

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$(x^{2}+2x)+(y^{2}-12y)=9$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

$(x^{2}+2x+1)+(y^{2}-12y+36)=9+1+36$

$(x^{2}+2x+1)+(y^{2}-12y+36)=46$

Rewrite as perfect squares

$(x+1)^{2}+(y-6)^{2}=46$    

therefore

the answer case F) is

$x^{2} +y^{2}+2x-12y-9=0$  ----->   $(x+1)^{2}+(y-6)^{2}=46$