When dealing with a trinomial of the form:
[tex]x^2+bx+c[/tex]
Compare the expression to that of a perfect square trinomial:
[tex](x+a)^2=x^2+2ax+a^2[/tex]
We want 2a=b, so we can add a^2 - a^2 to be able to write the trinomial as a squared binomial plus a constant. Since a=b/2, we should add b^2/4 - b^2/4.
C)
[tex]2x^2+2y^2+3x-5y=2[/tex]
First, notice that the coefficient of x^2 and y^2 is not 1. We need it to be equal to 1 in order to use the former procedure. Divide both sides by 2:
[tex]\begin{gathered} \Rightarrow\frac{2x^2+2y^2+3x-5y}{2}=\frac{2}{2} \\ \Rightarrow x^2+y^2+\frac{3}{2}x-\frac{5}{2}y=1 \end{gathered}[/tex]
The coefficient of x is 3/2, and comparing to the formula of the perfect square trinomial, then:
[tex]\begin{gathered} 2a=\frac{3}{2} \\ \Rightarrow a=\frac{3}{4} \\ \Rightarrow a^2=\frac{9}{16} \end{gathered}[/tex]
Te coefficient of y is -5/2, and comparing to the formula of the perfect square trinomial, the constant term inside the binomial should be -5/4, so we need a constant term of 25/15 to complete the square. Add 9/16 and 25/16 to both sides of the equation:
[tex]\Rightarrow x^2+y^2+\frac{3}{2}x-\frac{5}{2}y+\frac{9}{16}+\frac{25}{16}=1+\frac{9}{16}+\frac{25}{16}[/tex]
Reorder the terms on the left hand side of the equation, and simplify the expression on the right hand side:
[tex]\begin{gathered} \Rightarrow x^2+\frac{3}{2}x+\frac{9}{16}+y^2-\frac{5}{2}y+\frac{25}{16}=\frac{25}{8} \\ \Rightarrow(x+\frac{3}{4})^2+(y-\frac{5}{4})^2=\frac{25}{8} \end{gathered}[/tex]
The coordinates of the center of this circle, are:
[tex](-\frac{3}{4},\frac{5}{4})[/tex]
The radius is given by the square root of the constant term on the right hand side of the equation:
[tex]\begin{gathered} r=\frac{\sqrt[]{25}}{8} \\ \Rightarrow r=\frac{5}{4}\sqrt[]{2} \end{gathered}[/tex]
D)
[tex]x^2+y^2-2x-8y=8[/tex]
The coefficient of x is -2, which should be twice the constant term inside the squared binomial. So, that constant term is -1, and (-1)^2=1. Add 1 to both sides and rewrite the polynomial on x as a squared binomial:
[tex]\begin{gathered} \Rightarrow x^2+y^2-2x-8y+1=8+1 \\ \Rightarrow x^2-2x+1+y^2-8y=9 \\ \Rightarrow(x-1)^2+y^2-8y=9 \end{gathered}[/tex]
The coefficient of y is -8, which should be twice the constant term inside the squared binomial. So, the constant term is -4, and (-4)^2=16. Add 16 to both sides and rewrite the polynomial on y as a squared binomial:
[tex]\begin{gathered} \Rightarrow(x-1)^2+y^2-8y+16=9+16 \\ \Rightarrow(x-1)^2+(y-4)^2=25 \end{gathered}[/tex]
The coordinates of the center of this circumference are (1,4), and the radius is equal to 5.
