Two circles have an external tangent with length 36 cm. The shortest distance between these circles is 14 cm. If the radius of the longer circle is 4 times the radius of the smaller circle then the radius of the larger circle in cms is

Options

A. 12
B. 20
C. 14
D. 18

Answer

To solve the problem step by step, we will use the given information about the two circles and their relationship.

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

According to the problem, we know that:
\[ r_1 = 4r_2 \]

### Step 2: Understand the Geometry
The external tangent length between the two circles is given as 36 cm, and the shortest distance between the circles is 14 cm. The distance between the centers of the circles can be expressed as:
\[ d = r_1 + r_2 + 14 \]

### Step 3: Apply the Pythagorean Theorem
Using the right triangle formed by the centers of the circles and the tangent line, we can apply the Pythagorean theorem:
\[ d^2 = (r_1 - r_2)^2 + (36)^2 \]
Substituting \( d \) from the previous step:
\[ (r_1 + r_2 + 14)^2 = (r_1 - r_2)^2 + 36^2 \]

### Step 4: Substitute \( r_1 \)
Now, substitute \( r_1 = 4r_2 \) into the equation:
\[ (4r_2 + r_2 + 14)^2 = (4r_2 - r_2)^2 + 36^2 \]
This simplifies to:
\[ (5r_2 + 14)^2 = (3r_2)^2 + 36^2 \]

### Step 5: Expand Both Sides
Expanding both sides:
\[ (5r_2 + 14)^2 = 25r_2^2 + 140r_2 + 196 \]
And for the right side:
\[ (3r_2)^2 + 36^2 = 9r_2^2 + 1296 \]

### Step 6: Set Up the Equation
Now we can set up the equation:
\[ 25r_2^2 + 140r_2 + 196 = 9r_2^2 + 1296 \]
Rearranging gives:
\[ 25r_2^2 - 9r_2^2 + 140r_2 + 196 - 1296 = 0 \]
This simplifies to:
\[ 16r_2^2 + 140r_2 - 1100 = 0 \]

### Step 7: Simplify the Quadratic Equation
Dividing the entire equation by 4 gives:
\[ 4r_2^2 + 35r_2 - 275 = 0 \]

### Step 8: Solve the Quadratic Equation
Using the quadratic formula \( r_2 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 4 \), \( b = 35 \), and \( c = -275 \):
\[ r_2 = \frac{-35 \pm \sqrt{35^2 - 4 \cdot 4 \cdot (-275)}}{2 \cdot 4} \]
Calculating the discriminant:
\[ 35^2 = 1225 \]
\[ 4 \cdot 4 \cdot 275 = 4400 \]
So,
\[ r_2 = \frac{-35 \pm \sqrt{1225 + 4400}}{8} \]
\[ r_2 = \frac{-35 \pm \sqrt{5625}}{8} \]
\[ \sqrt{5625} = 75 \]
Thus,
\[ r_2 = \frac{-35 + 75}{8} \quad \text{(taking the positive root)} \]
\[ r_2 = \frac{40}{8} = 5 \text{ cm} \]

### Step 9: Find \( r_1 \)
Now, substituting back to find \( r_1 \):
\[ r_1 = 4r_2 = 4 \times 5 = 20 \text{ cm} \]

### Final Answer
The radius of the larger circle is \( \boxed{20} \) cm.