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 a degenerate case, give the solution set.
(x^2 + y^2 - 2x - 12y + 41 = 0)
part: 0 / 2
part 1 of 2
the equation in standard form is (square).
the equation represents (\text{select})

Explanation:

Step1: Group x and y terms

$x^2 - 2x + y^2 - 12y = -41$

Step2: Complete the square for x

Take half of -2, square it: $(-1)^2=1$. Add to both sides.
$x^2 - 2x + 1 + y^2 - 12y = -41 + 1$

Step3: Complete the square for y

Take half of -12, square it: $(-6)^2=36$. Add to both sides.
$(x^2 - 2x + 1) + (y^2 - 12y + 36) = -41 + 1 + 36$

Step4: Rewrite as perfect squares

$(x-1)^2 + (y-6)^2 = -4$

Answer:

The equation in standard form is $\boldsymbol{(x-1)^2 + (y-6)^2 = -4}$.
The equation represents a degenerate case with no real solution set (since the right-hand side is negative, there are no real points $(x,y)$ that satisfy the equation).