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 area of `C_(2)` is __________.

Options

A. `8 cm^(2)`
B. `8 sqrt(pi) cm^(2)`
C. `16 cm^(2)`
D. `16 sqrt(pi)cm^(2)`

Answer

To solve the problem step by step, we will follow these instructions:

### Step 1: Understand the given information
We have two circles, \( C_1 \) and \( C_2 \). Circle \( C_1 \) passes through the center of circle \( C_2 \) and is tangential to it. The area of circle \( C_1 \) is given as \( 4 \, \text{cm}^2 \).

### Step 2: Use the area formula to find the radius of circle \( C_1 \)
The area \( A \) of a circle is given by the formula:
\[ A = \pi r^2 \]
For circle \( C_1 \):
\[ 4 = \pi r_1^2 \]
where \( r_1 \) is the radius of circle \( C_1 \).

### Step 3: Solve for \( r_1 \)
Rearranging the equation gives:
\[ r_1^2 = \frac{4}{\pi} \]
Taking the square root of both sides:
\[ r_1 = \frac{2}{\sqrt{\pi}} \]

### Step 4: Determine the relationship between the radii of circles \( C_1 \) and \( C_2 \)
Since circle \( C_1 \) passes through the center of circle \( C_2 \) and is tangential to it, the radius \( r_2 \) of circle \( C_2 \) is equal to the diameter of circle \( C_1 \):
\[ r_2 = 2r_1 \]

### Step 5: Substitute \( r_1 \) into the equation for \( r_2 \)
Substituting the value of \( r_1 \):
\[ r_2 = 2 \left(\frac{2}{\sqrt{\pi}}\right) = \frac{4}{\sqrt{\pi}} \]

### Step 6: Calculate the area of circle \( C_2 \)
Using the area formula for circle \( C_2 \):
\[ A_2 = \pi r_2^2 \]
Substituting \( r_2 \):
\[ A_2 = \pi \left(\frac{4}{\sqrt{\pi}}\right)^2 \]
Calculating \( r_2^2 \):
\[ r_2^2 = \frac{16}{\pi} \]
Thus,
\[ A_2 = \pi \cdot \frac{16}{\pi} = 16 \, \text{cm}^2 \]

### Final Answer
The area of circle \( C_2 \) is \( 16 \, \text{cm}^2 \).

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