Two circles touch each externally. The sum of their areas is `130 pi` sq cm and the distance between their centres is 14 cm. The radius of the smaller circle is

Options

A. 2 cm
B. 4 cm
C. 5 cm
D. 3 cm

Answer

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

### Step 1: Define the variables
Let the radius of the larger circle be \( r_1 \) and the radius of the smaller circle be \( r_2 \).

### Step 2: Set up the equations
Since the two circles touch each other externally, the distance between their centers is equal to the sum of their radii:
\[ r_1 + r_2 = 14 \quad \text{(1)} \]
The sum of their areas is given as \( 130\pi \):
\[ \pi r_1^2 + \pi r_2^2 = 130\pi \]
Dividing through by \( \pi \) gives:
\[ r_1^2 + r_2^2 = 130 \quad \text{(2)} \]

### Step 3: Substitute \( r_1 \) in terms of \( r_2 \)
From equation (1), we can express \( r_1 \) in terms of \( r_2 \):
\[ r_1 = 14 - r_2 \]

### Step 4: Substitute into the area equation
Substituting \( r_1 \) into equation (2):
\[ (14 - r_2)^2 + r_2^2 = 130 \]

### Step 5: Expand and simplify
Expanding \( (14 - r_2)^2 \):
\[ 196 - 28r_2 + r_2^2 + r_2^2 = 130 \]
Combine like terms:
\[ 2r_2^2 - 28r_2 + 196 = 130 \]
Subtract \( 130 \) from both sides:
\[ 2r_2^2 - 28r_2 + 66 = 0 \]

### Step 6: Simplify the quadratic equation
Divide the entire equation by 2:
\[ r_2^2 - 14r_2 + 33 = 0 \]

### Step 7: Factor the quadratic equation
To factor the quadratic equation, we look for two numbers that multiply to \( 33 \) and add to \( -14 \). The factors are \( -11 \) and \( -3 \):
\[ (r_2 - 11)(r_2 - 3) = 0 \]

### Step 8: Solve for \( r_2 \)
Setting each factor to zero gives:
\[ r_2 - 11 = 0 \quad \Rightarrow \quad r_2 = 11 \]
\[ r_2 - 3 = 0 \quad \Rightarrow \quad r_2 = 3 \]

### Step 9: Determine the smaller radius
Since we are asked for the radius of the smaller circle, we take:
\[ r_2 = 3 \text{ cm} \]

### Final Answer
The radius of the smaller circle is \( 3 \) cm.

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