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: Recall circle - equation formula

The standard form of the equation of a circle is $(x - h)^2+(y - k)^2=r^2$, where $(h,k)$ is the center of the circle and $r$ is the radius. Since the center lies on the $x -$axis, $k = 0$. Given that the diameter $d = 12$, the radius $r=\frac{d}{2}=6$.

Step2: Analyze each option

• For the equation $(x - 12)^2+y^2 = 12$, here $r^2=12$, so $r=\sqrt{12}

eq6$.

• For the equation $(x - 6)^2+y^2 = 36$, here $h = 6,k = 0,r^2 = 36$, so $r = 6$.
• For the equation $x^2+y^2 = 12$, here $h = 0,k = 0,r^2 = 12$, so $r=\sqrt{12}

eq6$.

• For the equation $x^2+y^2 = 144$, here $h = 0,k = 0,r^2 = 144$, so $r = 12

eq6$.

• For the equation $(x + 6)^2+y^2 = 36$, here $h=-6,k = 0,r^2 = 36$, so $r = 6$.
• For the equation $(x + 12)^2+y^2 = 144$, here $h=-12,k = 0,r^2 = 144$, so $r = 12

eq6$.

Answer:

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