Answer:
[tex][/tex]Explanation:
Given:
[tex]4x^2+6y=-4y^2+12[/tex]To find:
convert from general to standard form
The standard form of a circle:
[tex](x\text{ - a\rparen}^2\text{ + \lparen y - b\rparen}^2\text{ = r}^2[/tex]To convert the given form to the standard form, we will write x and y into perfect squares using the complete the square:
[tex]\begin{gathered} first\text{ divide through by 4 \lparen we need the coefficient of x}^2\text{ and y}^2\text{ to be 1\rparen} \\ \frac{4x^2}{4}\text{ + }\frac{6y}{4}\text{ = }\frac{-4y^2}{4}\text{ + }\frac{12}{4} \\ \\ x^2\text{ + }\frac{3}{2}y\text{ = -y}^2\text{ + 3} \\ \\ x^2\text{ + y}^2\text{ +}\frac{3}{2}y\text{ = 3} \end{gathered}[/tex][tex]\begin{gathered} for\text{ x, we only have x}^2.\text{ There is no coefficient of x. We can't complete the square} \\ \\ for\text{ y: }y^2\text{ + }\frac{3}{2}y \\ We\text{ will complete the square:} \\ coefficient\text{ of y = 3/2; half the coefficient of y = 3/4} \\ square\text{ of half the coefficient of y = \lparen}\frac{3}{4})\placeholder{⬚}^2 \\ \\ Add\text{ square of half the coefficient of y to both sides of the equation:} \\ x^2\text{ + y}^2\text{ + }\frac{3}{2}y\text{ + \lparen}\frac{3}{4})\placeholder{⬚}^2\text{ = 3 + \lparen}\frac{3}{4})\placeholder{⬚}^2 \\ \\ x^2\text{ + \lparen y + }\frac{3}{4})\placeholder{⬚}^2\text{ = 3 + }\frac{9}{16} \end{gathered}[/tex][tex]\begin{gathered} x^2\text{ + \lparen y + }\frac{3}{4})\placeholder{⬚}^2\text{ = }\frac{3(16)\text{ + 9}}{16} \\ \\ x^2\text{ + \lparen y + }\frac{3}{4})\placeholder{⬚}^2\text{ = }\frac{57}{16} \end{gathered}[/tex]
