Question

a circle has a diameter of 12 units, and its center lies on the x-axis. what could be the equation of the circle? check all that apply.$(x-12)^2+y^2=12$$(x-6)^2+y^2=36$$x^2+y^2=12$$x^2+y^2=144$$(x+6)^2+y^2=36$$(x+12)^2+y^2=144$

Explanation:

Step1: Find radius from diameter

The diameter is 12, so radius $r = \frac{12}{2}=6$. The radius squared is $r^2=6^2=36$.

Step2: Check center on x-axis

The standard circle equation is $(x-h)^2+(y-k)^2=r^2$. For center on x-axis, $k=0$, so equation is $(x-h)^2+y^2=36$, where $h$ is any real number.

Step3: Verify each option

1. $(x-12)^2+y^2=12$: $r^2=12

eq36$, invalid.

1. $(x-6)^2+y^2=36$: $r^2=36$, $k=0$, valid.
2. $x^2+y^2=12$: $r^2=12

eq36$, invalid.

1. $x^2+y^2=144$: $r^2=144

eq36$, invalid.

1. $(x+6)^2+y^2=36$: $r^2=36$, $k=0$, valid.
2. $(x+12)^2+y^2=144$: $r^2=144

eq36$, invalid.

Answer:

B. $(x-6)^2+y^2=36$, E. $(x+6)^2+y^2=36$