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
Let the centers of the two circles be \( O_1 \) and \( O_2 \). The distance between the centers \( O_1O_2 \) is given as 25 units. The radius of the first circle (centered at \( O_1 \)) is 15 units, and the radius of the second circle (centered at \( O_2 \)) is 20 units.

### Step 2: Set Up the Right Triangle
Let the common chord intersect the line segment \( O_1O_2 \) at point \( P \). Drop perpendiculars from \( O_1 \) and \( O_2 \) to the chord at point \( P \). Let \( O_1P = h \) (the distance from the center of the first circle to the chord) and \( O_2P = h' \) (the distance from the center of the second circle to the chord).

### Step 3: Use the Pythagorean Theorem
In triangle \( O_1AP \):
\[ O_1A^2 + h^2 = O_1P^2 \]
Where \( O_1A = 15 \) (radius of the first circle), so:
\[ 15^2 = h^2 + x^2 \quad (1) \]

In triangle \( O_2BP \):
\[ O_2B^2 + h'^2 = O_2P^2 \]
Where \( O_2B = 20 \) (radius of the second circle), so:
\[ 20^2 = h'^2 + (25 - x)^2 \quad (2) \]

### Step 4: Express \( h' \) in terms of \( h \)
Since the total distance between the centers is 25 units:
\[ h + h' = 25 \]
From this, we can express \( h' \) as:
\[ h' = 25 - h \]

### Step 5: Substitute and Solve
Substituting \( h' \) into equation (2):
\[ 20^2 = (25 - h)^2 + x^2 \]
Expanding this:
\[ 400 = (625 - 50h + h^2) + x^2 \]
Rearranging gives:
\[ h^2 + x^2 - 50h + 225 = 0 \quad (3) \]

### Step 6: Solve Equations (1) and (3)
From equation (1):
\[ h^2 + x^2 = 225 \]
Substituting into equation (3):
\[ 225 - 50h + 225 = 0 \]
This simplifies to:
\[ -50h + 450 = 0 \implies h = 9 \]

### Step 7: Find \( x \)
Substituting \( h = 9 \) back into equation (1):
\[ 15^2 = 9^2 + x^2 \implies 225 = 81 + x^2 \implies x^2 = 144 \implies x = 12 \]

### Step 8: Length of the Common Chord
The length of the common chord \( AB \) is twice the value of \( x \):
\[ AB = 2x = 2 \times 12 = 24 \]

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

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