Question

which equations represent circles that have a diameter of 12 units and a center that lies on the y - axis? select two options.
$x^{2}+(y - 3)^{2}=36$
$x^{2}+(y - 5)^{2}=6$
$(x - 4)^{2}+y^{2}=36$
$(x + 6)^{2}+y^{2}=144$
$x^{2}+(y + 8)^{2}=36$

Explanation:

Step1: Recall circle - equation form

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

Step2: Analyze each option

Option 1: $x^{2}+(y - 3)^{2}=36$

Here, $h = 0,k = 3,r^{2}=36$, so $r = 6$. The center $(0,3)$ lies on the $y$ - axis.

Option 2: $x^{2}+(y - 5)^{2}=6$

Here, $r^{2}=6$, so $r=\sqrt{6}
eq6$.

Option 3: $(x - 4)^{2}+y^{2}=36$

Here, $h = 4
eq0$, so the center $(4,0)$ does not lie on the $y$ - axis.

Option 4: $(x + 6)^{2}+y^{2}=144$

Here, $r^{2}=144$, so $r = 12
eq6$, and $h=-6
eq0$.

Option 5: $x^{2}+(y + 8)^{2}=36$

Here, $h = 0,k=-8,r^{2}=36$, so $r = 6$. The center $(0,-8)$ lies on the $y$ - axis.

Answer:

$x^{2}+(y - 3)^{2}=36$, $x^{2}+(y + 8)^{2}=36$