Two circles with centres O and O' of radii 3 cm and 4 cm , respectively intersect at two points P and Q such that OP and O'P are tangents to the two circle . Find the length of the common chord PQ.

Answer

To find the length of the common chord \( PQ \) of the two intersecting circles with centers \( O \) and \( O' \) and radii \( 3 \, \text{cm} \) and \( 4 \, \text{cm} \) respectively, we can follow these steps:

### Step 1: Understand the Geometry
We have two circles:
- Circle with center \( O \) and radius \( r_1 = 3 \, \text{cm} \)
- Circle with center \( O' \) and radius \( r_2 = 4 \, \text{cm} \)

The points \( P \) and \( Q \) are the points of intersection of the two circles. The lines \( OP \) and \( O'P \) are tangents to the circles at point \( P \).

### Step 2: Use the Right Triangle Property
Since \( OP \) and \( O'P \) are tangents to the circles, we know that:
- \( OP \perp OQ \)
- \( O'P \perp O'Q \)

Thus, triangles \( OPO' \) and \( OQO' \) are right triangles.

### Step 3: Apply the Pythagorean Theorem
Let \( OO' \) be the distance between the centers of the circles. According to the Pythagorean theorem in triangle \( OPO' \):
\[ OO'^2 = OP^2 + O'P^2 \]
Substituting the known values:
\[ OO'^2 = 3^2 + 4^2 = 9 + 16 = 25 \]
Taking the square root:
\[ OO' = \sqrt{25} = 5 \, \text{cm} \]

### Step 4: Find the Length of the Common Chord
Let \( M \) be the midpoint of the common chord \( PQ \). The radius \( OM \) is perpendicular to the chord \( PQ \).

Using the right triangle \( OMP \):
\[ OP^2 = OM^2 + PM^2 \]
Where \( PM \) is half the length of the chord \( PQ \) (denote it as \( x \)):
\[ 3^2 = OM^2 + x^2 \]
\[ 9 = OM^2 + x^2 \quad \text{(1)} \]

Similarly, for the circle with center \( O' \):
\[ O'P^2 = O'M^2 + PM^2 \]
\[ 4^2 = O'M^2 + x^2 \]
\[ 16 = O'M^2 + x^2 \quad \text{(2)} \]

### Step 5: Express \( OM \) and \( O'M \)
From the triangle \( OOO' \):
\[ OM + O'M = OO' = 5 \]
Let \( OM = y \) and \( O'M = 5 - y \).

Substituting into equations (1) and (2):
1. \( 9 = y^2 + x^2 \)
2. \( 16 = (5 - y)^2 + x^2 \)

### Step 6: Solve the Equations
From equation (1):
\[ x^2 = 9 - y^2 \quad \text{(3)} \]

Substituting (3) into equation (2):
\[ 16 = (5 - y)^2 + (9 - y^2) \]
Expanding:
\[ 16 = 25 - 10y + y^2 + 9 - y^2 \]
\[ 16 = 34 - 10y \]
\[ 10y = 34 - 16 \]
\[ 10y = 18 \quad \Rightarrow \quad y = 1.8 \]

### Step 7: Find \( x \)
Substituting \( y = 1.8 \) back into equation (3):
\[ x^2 = 9 - (1.8)^2 = 9 - 3.24 = 5.76 \]
Taking the square root:
\[ x = \sqrt{5.76} = 2.4 \]

### Step 8: Find the Length of the Common Chord \( PQ \)
Since \( PQ = 2 \times PM = 2x \):
\[ PQ = 2 \times 2.4 = 4.8 \, \text{cm} \]

### Final Answer
The length of the common chord \( PQ \) is \( 4.8 \, \text{cm} \).

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