Two circles touch externally. The sum of their areas is `130pi`. 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 find the radius of the bigger circle, we can follow these steps:

### Step 1: Define the variables
Let:
- \( r_1 \) = radius of the bigger circle
- \( r_2 \) = radius of the smaller circle

### Step 2: Set up the equations
From the problem, we know:
1. The sum of the areas of the circles is \( 130\pi \) square cm:
\[ \pi r_1^2 + \pi r_2^2 = 130\pi \]
Dividing both sides by \( \pi \):
\[ r_1^2 + r_2^2 = 130 \quad \text{(Equation 1)} \]

2. The distance between the centers of the circles is \( 14 \) cm:
\[ r_1 + r_2 = 14 \quad \text{(Equation 2)} \]

### Step 3: Express \( r_2 \) in terms of \( r_1 \)
From Equation 2, we can express \( r_2 \):
\[ r_2 = 14 - r_1 \]

### Step 4: Substitute \( r_2 \) into Equation 1
Substituting \( r_2 \) in Equation 1:
\[ r_1^2 + (14 - r_1)^2 = 130 \]

### Step 5: Expand and simplify the equation
Expanding \( (14 - r_1)^2 \):
\[ r_1^2 + (196 - 28r_1 + r_1^2) = 130 \]
Combining like terms:
\[ 2r_1^2 - 28r_1 + 196 = 130 \]
Subtracting \( 130 \) from both sides:
\[ 2r_1^2 - 28r_1 + 66 = 0 \]

### Step 6: Simplify the quadratic equation
Dividing the entire equation by \( 2 \):
\[ r_1^2 - 14r_1 + 33 = 0 \]

### Step 7: Factor the quadratic equation
We need to factor \( r_1^2 - 14r_1 + 33 \):
\[ (r_1 - 11)(r_1 - 3) = 0 \]

### Step 8: Solve for \( r_1 \)
Setting each factor to zero gives us:
1. \( r_1 - 11 = 0 \) → \( r_1 = 11 \)
2. \( r_1 - 3 = 0 \) → \( r_1 = 3 \)

### Step 9: Identify the bigger radius
Since \( r_1 \) is the radius of the bigger circle, we take:
\[ r_1 = 11 \text{ cm} \]

### Conclusion
The radius of the bigger circle is \( 11 \) cm.

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