Question

identify this conic section. $x^2 - 4x + y^2 - 4y + 4 = 12$○ hyperbola○ ellipse○ circle○ line○ parabola

Explanation:

Step1: Rearrange given equation

$x^2 - 4x + y^2 - 4y = 12 - 4$
$x^2 - 4x + y^2 - 4y = 8$

Step2: Complete the square for x

Take half of -4, square it: $(-2)^2=4$. Add to both sides.
$x^2 - 4x + 4 + y^2 - 4y = 8 + 4$
$(x-2)^2 + y^2 - 4y = 12$

Step3: Complete the square for y

Take half of -4, square it: $(-2)^2=4$. Add to both sides.
$(x-2)^2 + y^2 - 4y + 4 = 12 + 4$
$(x-2)^2 + (y-2)^2 = 16$

Step4: Match to conic form

The equation is in the standard circle form $(x-h)^2 + (y-k)^2 = r^2$, where $(h,k)$ is the center and $r$ is the radius.

Answer:

circle