Answer:
A: 2√(12/13)
Step-by-step explanation:
Let R be the radius of the larger circle.
Let r be the radius of the smaller circle.
The area of the larger circle with radius R = 2 is:
[tex]A_{\sf large\;circle}=\pi \cdot R^2 \\\\ A_{\sf large\;circle}=\pi \cdot 2^2 \\\\ A_{\sf large\;circle}=4\pi[/tex]
The area of the smaller circle with radius r is:
[tex]A_{\sf small\;circle}=\pi r^2[/tex]
When the smaller circle is cut out of the larger circle, the remaining area can be expressed as:
[tex]A_{\sf remaining}=A_{\sf large\;circle} -A_{\sf small\;circle} \\\\A_{\sf remaining}=4\pi -\pi r^2[/tex]
Given that the area of the smaller circle is 12 times the area of the remaining part, then:
[tex]A_{\sf small\;circle}=12 \cdot A_{\sf remaining} \\\\\pi r^2 = 12 \cdot (4\pi -\pi r^2)[/tex]
Solve for the radius of the smaller circle (r):
[tex]\pi r^2 = 12 \cdot (4\pi -\pi r^2) \\\\ \pi r^2 =48\pi -12\pi r^2 \\\\ \pi r^2 =\pi (48-12 r^2) \\\\ r^2=48-12r^2 \\\\r^2+12r^2=48\\\\13r^2=48\\\\r^2=\dfrac{48}{13}\\\\\\r=\sqrt{\dfrac{48}{13}}\\\\\\r=\sqrt{4\cdot \dfrac{12}{13}}\\\\\\r=\sqrt{4}\sqrt{\dfrac{12}{13}}\\\\\\r=2\sqrt{\dfrac{12}{13}}[/tex]
Therefore, the radius of the smaller circle is:
[tex]\Large\boxed{\boxed{2\sqrt{\dfrac{12}{13}}}}[/tex]
