The length of the common chord of two circles of radii 15 and 20, whose centres are 25 units apart, is

Options

A. 24
B. 25
C. 15
D. 20

Explanation

To find the length of the common chord of two circles with radii 15 and 20 units, whose centers are 25 units apart, we can follow these steps:

### Step 1: Understand the Geometry
We have two circles:
- Circle 1 with center O and radius 15 units.
- Circle 2 with center P and radius 20 units.
The distance between the centers O and P is 25 units.

### Step 2: Set Up the Problem
Let the common chord be PQ. The line segment OP (the distance between the centers) bisects the common chord PQ at point M. Therefore, we can denote:
- OM = x (the distance from O to the midpoint M of the chord)
- PM = 25 - x (the distance from P to M)

### Step 3: Apply the Pythagorean Theorem
In triangle OMQ:
- OM is perpendicular to PQ.
- By the Pythagorean theorem:
\[ OQ^2 = OM^2 + MQ^2 \]
Since OQ is the radius of the first circle (15 units), we have:
\[ 15^2 = x^2 + \left(\frac{D}{2}\right)^2 \quad \text{(Equation 1)} \]

In triangle PMQ:
- Similarly, we apply the Pythagorean theorem:
\[ PM^2 = PM^2 + MQ^2 \]
Since PM is the radius of the second circle (20 units), we have:
\[ 20^2 = (25 - x)^2 + \left(\frac{D}{2}\right)^2 \quad \text{(Equation 2)} \]

### Step 4: Substitute Values
From Equation 1:
\[ 225 = x^2 + \left(\frac{D}{2}\right)^2 \quad \text{(1)} \]
From Equation 2:
\[ 400 = (25 - x)^2 + \left(\frac{D}{2}\right)^2 \quad \text{(2)} \]

### Step 5: Expand and Rearrange Equation 2
Expanding Equation 2:
\[ 400 = 625 - 50x + x^2 + \left(\frac{D}{2}\right)^2 \]
Rearranging gives:
\[ 400 - 625 = -50x + x^2 + \left(\frac{D}{2}\right)^2 \]
\[ -225 = -50x + x^2 + \left(\frac{D}{2}\right)^2 \]
Now, substituting \(\left(\frac{D}{2}\right)^2\) from Equation 1 into this equation.

### Step 6: Solve for x
Subtract Equation 1 from the rearranged Equation 2:
\[ -225 = -50x + x^2 + \left(\frac{D}{2}\right)^2 - (x^2 + \left(\frac{D}{2}\right)^2) \]
This simplifies to:
\[ -225 = -50x \]
Thus,
\[ 50x = 450 \implies x = 9 \]

### Step 7: Find D
Substituting \(x = 9\) back into Equation 1:
\[ 225 = 9^2 + \left(\frac{D}{2}\right)^2 \]
\[ 225 = 81 + \left(\frac{D}{2}\right)^2 \]
\[ \left(\frac{D}{2}\right)^2 = 144 \implies \frac{D}{2} = 12 \implies D = 24 \]

### Final Answer
The length of the common chord is **24 units**.

---