Answer:
Options 2 and 5.
Step-by-step explanation:
The standard form of a circle is
[tex](x-h)^2+(y-k)^2=r^2[/tex] ... (1)
where, (h,k) is center and r is radius.
We need to find the circle that has a diameter of 12 units, and its center lies on the x-axis.
[tex]radius=\dfrac{Diameter}{2}=\dfrac{12}{2}=6[/tex]
So, radius of required circle must be 6 and center is in the form of (a,0).
The first equation is
[tex](x-12)^2+(y)^2=12[/tex] .... (2)
On comparing (1) and (2) we get
[tex]h=12,k=0,r=\sqrt{12}[/tex]
Center of the circle is (12,0) and radius is [tex]\sqrt{12}[/tex]. So, option 1 is incorrect.
Similarly,
For equation 2, center of the circle is (6,0) and radius is [tex]6[/tex]. So, option 2 is correct.
For equation 3, center of the circle is (0,0) and radius is [tex]\sqrt{12}[/tex]. So, option 3 is incorrect.
For equation 4, center of the circle is (0,0) and radius is [tex]12[/tex]. So, option 4 is incorrect.
For equation 5, center of the circle is (-6,0) and radius is [tex]6[/tex]. So, option 5 is correct.
For equation 6, center of the circle is (-12,0) and radius is [tex]12[/tex]. So, option 6 is incorrect.
Therefore, the correct options are 2 and 5.
