Circle `C_1` passes through the centre of circle `C_2` and is tangential to it. If the area of `C_1` is `4 cm^2`, then the area of `C_2` is _______

Options

A. `8 cm^2`
B. `8sqrtpi cm^2`
C. `16 cm^2`
D. `16sqrtpi cm^2`

Answer

To solve the problem, we need to find the area of circle \( C_2 \) given that circle \( C_1 \) passes through the center of circle \( C_2 \) and is tangential to it. We also know that the area of circle \( C_1 \) is \( 4 \, \text{cm}^2 \).

### Step-by-Step Solution:

1. **Find the Radius of Circle \( C_1 \)**:
The area \( A \) of a circle is given by the formula:
\[ A = \pi r^2 \]
where \( r \) is the radius of the circle. Given that the area of circle \( C_1 \) is \( 4 \, \text{cm}^2 \), we can set up the equation:
\[ \pi r_1^2 = 4 \]
To find \( r_1 \), we rearrange the equation:
\[ r_1^2 = \frac{4}{\pi} \]
Taking the square root:
\[ r_1 = \sqrt{\frac{4}{\pi}} = \frac{2}{\sqrt{\pi}} \]

2. **Understanding the Relationship between the Two Circles**:
Circle \( C_1 \) passes through the center of circle \( C_2 \) and is tangential to it. This means that the distance from the center of circle \( C_2 \) to the edge of circle \( C_1 \) is equal to the radius of circle \( C_2 \) (denote it as \( r_2 \)).

The distance from the center of circle \( C_2 \) to the edge of circle \( C_1 \) is equal to the radius of circle \( C_1 \) plus the radius of circle \( C_2 \):
\[ r_1 + r_2 = r_2 \]
Rearranging gives:
\[ r_2 = r_1 \]

3. **Finding the Radius of Circle \( C_2 \)**:
Since we have established that \( r_2 = r_1 \), we substitute the value of \( r_1 \):
\[ r_2 = \frac{2}{\sqrt{\pi}} \]

4. **Calculate the Area of Circle \( C_2 \)**:
Now we can find the area of circle \( C_2 \) using the area formula:
\[ A_2 = \pi r_2^2 \]
Substituting \( r_2 \):
\[ A_2 = \pi \left(\frac{2}{\sqrt{\pi}}\right)^2 = \pi \cdot \frac{4}{\pi} = 4 \, \text{cm}^2 \]

Thus, the area of circle \( C_2 \) is \( 4 \, \text{cm}^2 \).

### Final Answer:
The area of \( C_2 \) is \( 4 \, \text{cm}^2 \).