a) We have to find the equation of the quadratic expression.
We have the expression:
[tex]x^2+2x+y^2-4y=4[/tex]We can grouop the terms for y and x and find the constants we need to form a perfect binomial (completing the squares):
[tex]\begin{gathered} (x^2+2x+1-1)+(y^2-4y+4-4)=4 \\ \lbrack(x+1)^2-1\rbrack+\lbrack(y-2)^2-4\rbrack=4 \\ (x+1)^2+(y-2)^2-1-4=4 \\ (x+1)^2+(y-2)^2=4+1+4 \\ (x+1)^2+(y-2)^2=9 \\ (x+1)^2+(y-2)^2=3^2 \end{gathered}[/tex]The expression results in a circumference of radius 3 with the center ar (-1,2).
c) If we have the equation:
[tex]2x^2+2y^2+3x-5y=2[/tex]We can solve this as:
[tex]\begin{gathered} 2(x^2+y^2+\frac{3}{2}x-\frac{5}{2}y)=2 \\ \lbrack x^2+\frac{3}{2}x+(\frac{3}{4})^2-(\frac{3}{4})^2\rbrack+\lbrack y^2-\frac{5}{2}y+(\frac{5}{4})^2-(\frac{5}{4})^2\rbrack=1 \\ \lbrack(x+\frac{3}{4})^2-\frac{9}{16}\rbrack+\lbrack(y-\frac{5}{4})^2-\frac{25}{16}=1 \\ (x+\frac{3}{4})^2+(y-\frac{5}{4})^2=1+\frac{9}{16}+\frac{25}{16} \\ (x+\frac{3}{4})^2+(y-\frac{5}{4})^2=\frac{16+9+25}{16} \\ (x+\frac{3}{4})^2+(y-\frac{5}{4})^2=\frac{50}{16} \end{gathered}[/tex]We can calculate the radius as:
[tex]r=\sqrt[]{\frac{50}{16}}=\sqrt[]{\frac{2\cdot25}{16}}=\sqrt[]{2}\cdot\frac{5}{4}=\frac{5}{4}\sqrt[]{2}[/tex]
