Select from the following two-variable quadratic equations those which are

The general equation of circle is ${\displaystyle x^2+y^2+2gx+2fy+c=0}$
center is ${\displaystyle (-g,-f)}$ and radius is ${\displaystyle \sqrt{8^2+f^2-c}}$
a) ${\displaystyle x^2+y^2+6x-4y+4=0}$
Comparing this equation with the general equation we get:
${\displaystyle 2g=6\Rightarrow g=3}$
${\displaystyle 2f=-4\Rightarrow f=-2}$
${\displaystyle R=\sqrt{3^2+{(-2)}^2-4}=3}$
Therefore the center is ${\displaystyle (-3,2)}$ and radius is 3.
b) ${\displaystyle x^2+y^2+2xy+5x-3y-5=0}$
This is not an equation of circle
c) ${\displaystyle x^2+y^2-10x+6y+35=0}$
This can be written as ${\displaystyle {(x-h)}^2+{(y-k)}^2=r^2}$
${\displaystyle {(x-5)}^2+{(y+3)}^2={5.57}^2}$
Therefore the center is ${\displaystyle (5,-3)}$ and radius is 5.57.
d) ${\displaystyle -2x^2-2y^2+4x+y+22=0}$
Divide the equation by -2
${\displaystyle x^2+y^2-2x-4y-11=0}$
From comparing the above equation with the general equation we get:
${\displaystyle 2g=-2\Rightarrow g=-1}$
${\displaystyle 2f=-4\Rightarrow f=-2}$
${\displaystyle R=\sqrt{1^2+2^2+11}=\sqrt{1+4+11}=4}$
Therefore, center is (1,2) and radius is 4.