Two circles touch externally. The sum of their areas is `130 pi` sq. cm. and the distance between their centres is 14 cm. The radius of the bigger circle is
(Take `pi = (22)/(7)`)

Options

A. 22 cm
B. 11 cm
C. 33 cm
D. 44 cm

Answer

To solve the problem step by step, we will denote the radius of the bigger circle as \( r_2 \) and the radius of the smaller circle as \( r_1 \).

### Step 1: Set up the equations based on the problem statement.
- We know that the two circles touch externally, so the distance between their centers is equal to the sum of their radii:
\[ r_1 + r_2 = 14 \quad \text{(Equation 1)} \]

- We also know that the sum of their areas is \( 130\pi \) square centimeters:
\[ \pi r_1^2 + \pi r_2^2 = 130\pi \]
Dividing the entire equation by \( \pi \):
\[ r_1^2 + r_2^2 = 130 \quad \text{(Equation 2)} \]

### Step 2: 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 3: Substitute \( r_1 \) into Equation 2.
Now, we substitute \( r_1 \) into Equation 2:
\[ (14 - r_2)^2 + r_2^2 = 130 \]

### Step 4: Expand and simplify the equation.
Expanding \( (14 - r_2)^2 \):
\[ 196 - 28r_2 + r_2^2 + r_2^2 = 130 \]
Combining like terms:
\[ 2r_2^2 - 28r_2 + 196 = 130 \]
Subtracting 130 from both sides:
\[ 2r_2^2 - 28r_2 + 66 = 0 \]

### Step 5: Simplify the quadratic equation.
Dividing the entire equation by 2:
\[ r_2^2 - 14r_2 + 33 = 0 \]

### Step 6: Factor the quadratic equation.
To factor the quadratic equation:
\[ (r_2 - 3)(r_2 - 11) = 0 \]
This gives us two possible solutions for \( r_2 \):
\[ r_2 = 3 \quad \text{or} \quad r_2 = 11 \]

### Step 7: Identify the radius of the bigger circle.
Since \( r_2 \) is the radius of the bigger circle, we take the larger value:
\[ r_2 = 11 \text{ cm} \]

### Final Answer:
The radius of the bigger circle is \( 11 \) cm.

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