Q:

# Match the circle equations in general form with their corresponding equations in standard form.Tilesx2 + y2− 4x + 12y − 20 = 0(x − 6)2+ (y − 4)2 = 56x2 + y2+ 6x − 8y − 10 = 0 (x − 2)2+ (y + 6)2 = 603x2 + 3y2 + 12x + 18y − 15 = 0(x + 2)2+ (y + 3)2 = 185x2 + 5y2 − 10x + 20y − 30 = 0(x + 1)2+ (y − 6)2 = 462x2 + 2y2 − 24x − 16y − 8 = 0x2 + y2+ 2x − 12y − 9 =

Accepted Solution

A:
case A) $$x^{2} +y^{2} -4x+12y-20=0$$Convert to standard formGroup 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 $$(x-2)^{2}+(y+6)^{2}=60$$thereforethe 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 $$(x+3)^{2}+(y-4)^{2}=35$$thereforethe 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 $$(x+2)^{2}+(y+3)^{2}=18$$thereforethe 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 $$(x-1)^{2}+(y+2)^{2}=11$$     thereforethe 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 $$(x-6)^{2}+(y-4)^{2}=56$$     thereforethe 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 $$(x+1)^{2}+(y-6)^{2}=46$$     thereforethe answer case F) is$$x^{2} +y^{2}+2x-12y-9=0$$  ----->   $$(x+1)^{2}+(y-6)^{2}=46$$