Question

write the equation in the form $(x - h)^2+(y - k)^2 = c$. then, if the equation represents a circle, identify the center and radius. if the equation represents the degenerate case, give the solution set.
$x^{2}+y^{2}-12x + 8y+3 = 0$

Explanation:

Step1: Group the x - and y - terms

$x^{2}-12x+y^{2}+8y+3 = 0$

Step2: Complete the square for the x - terms

$x^{2}-12x=(x - 6)^{2}-36$.

Step3: Complete the square for the y - terms

$y^{2}+8y=(y + 4)^{2}-16$.

Step4: Substitute into the original equation

$(x - 6)^{2}-36+(y + 4)^{2}-16+3=0$.

Step5: Simplify the equation

$(x - 6)^{2}+(y + 4)^{2}=36 + 16-3$.
$(x - 6)^{2}+(y + 4)^{2}=49$.
For a circle equation $(x - h)^2+(y - k)^2=r^{2}$, here $h = 6$, $k=-4$ and $r^{2}=49$ (so $r = 7$).

Answer:

The equation in the form $(x - h)^2+(y - k)^2= c$ is $(x - 6)^2+(y + 4)^2= 49$. The center of the circle is $(6,-4)$ and the radius is $7$.