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^2-6x - 12y + 41 = 0

Explanation:

Step1: Group x - terms and y - terms

$(x^{2}-6x)+(y^{2}-12y)+41 = 0$

Step2: Complete the square for x - terms

$x^{2}-6x=(x - 3)^{2}-9$. So the equation becomes $(x - 3)^{2}-9+(y^{2}-12y)+41 = 0$.

Step3: Complete the square for y - terms

$y^{2}-12y=(y - 6)^{2}-36$. Then the equation is $(x - 3)^{2}-9+(y - 6)^{2}-36 + 41=0$.

Step4: Simplify the equation

$(x - 3)^{2}+(y - 6)^{2}-9-36 + 41=0$, which simplifies to $(x - 3)^{2}+(y - 6)^{2}=4$.
Since the equation is in the form $(x - h)^2+(y - k)^2=r^{2}$ (where $h = 3,k = 6,r = 2$), the center of the circle is $(h,k)=(3,6)$ and the radius $r = 2$.

Answer:

The equation in the form $(x - h)^2+(y - k)^2=c$ is $(x - 3)^2+(y - 6)^2 = 4$. The center is $(3,6)$ and the radius is $2$.