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

Select from the following two-variable quadratic equations those which are equations of circles. Determine the center and the radius of the circles.
a. ${x}^{2}+{y}^{2}+6x-4y+4=0$
b. ${x}^{2}+{y}^{2}+2xy+5x-3y-5=0$
c. ${x}^{2}+{y}^{2}-10x+6y+35=0$
d. $-2{x}^{2}-2{y}^{2}+4x+8y+22=0$
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junoon363km
The general equation of circle is ${x}^{2}+{y}^{2}+2gx+2fy+c=0$
center is $\left(-g,-f\right)$ and radius is $\sqrt{{8}^{2}+{f}^{2}-c}$
a) ${x}^{2}+{y}^{2}+6x-4y+4=0$
Comparing this equation with the general equation we get:
$2g=6⇒g=3$
$2f=-4⇒f=-2$
$R=\sqrt{{3}^{2}+{\left(-2\right)}^{2}-4}=3$
Therefore the center is $\left(-3,2\right)$ and radius is 3.
b) ${x}^{2}+{y}^{2}+2xy+5x-3y-5=0$
This is not an equation of circle
c) ${x}^{2}+{y}^{2}-10x+6y+35=0$
This can be written as ${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={r}^{2}$
${\left(x-5\right)}^{2}+{\left(y+3\right)}^{2}={5.57}^{2}$
Therefore the center is $\left(5,-3\right)$ and radius is 5.57.
d) $-2{x}^{2}-2{y}^{2}+4x+y+22=0$
Divide the equation by -2
${x}^{2}+{y}^{2}-2x-4y-11=0$
From comparing the above equation with the general equation we get:
$2g=-2⇒g=-1$
$2f=-4⇒f=-2$
$R=\sqrt{{1}^{2}+{2}^{2}+11}=\sqrt{1+4+11}=4$
Therefore, center is (1,2) and radius is 4.