AC is transverse common tangent to two circles with centres P and Q and radii 6 cm and 3 cm at the point A and C respectively. If AC cuts PQ at the point B and AB= 8cm then the length of `PQ` is:

Options

A. 13 cm
B. 12 cm
C. 10 cm
D. 15 cm

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Geometry
We have two circles with centers P and Q. The radius of the circle at P is 6 cm, and the radius of the circle at Q is 3 cm. The line AC is a transverse common tangent to both circles, touching them at points A and C respectively.

### Step 2: Identify Given Information
- Radius of circle at P (PA) = 6 cm
- Radius of circle at Q (CQ) = 3 cm
- Length AB = 8 cm (where B is the point where AC intersects PQ)

### Step 3: Use the Right Triangle
In triangle PAB, we can apply the Pythagorean theorem since PA is perpendicular to AC at point A.

Using Pythagorean theorem:
\[ PB^2 = PA^2 + AB^2 \]
Substituting the known values:
\[ PB^2 = 6^2 + 8^2 = 36 + 64 = 100 \]
Taking the square root:
\[ PB = \sqrt{100} = 10 \text{ cm} \]

### Step 4: Establish Similar Triangles
Triangles PAB and BCQ are similar because they share angle B and both have a right angle. Therefore, we can set up a proportion:
\[ \frac{BC}{CQ} = \frac{AB}{PA} \]
Substituting the known values:
\[ \frac{BC}{3} = \frac{8}{6} \]
Cross-multiplying gives:
\[ BC \cdot 6 = 8 \cdot 3 \implies 6BC = 24 \implies BC = 4 \text{ cm} \]

### Step 5: Apply Pythagorean Theorem in Triangle BCQ
Now we will use triangle BCQ to find BQ:
\[ BQ^2 = BC^2 + CQ^2 \]
Substituting the known values:
\[ BQ^2 = 4^2 + 3^2 = 16 + 9 = 25 \]
Taking the square root:
\[ BQ = \sqrt{25} = 5 \text{ cm} \]

### Step 6: Calculate PQ
Finally, we can find the length of PQ:
\[ PQ = PB + BQ = 10 + 5 = 15 \text{ cm} \]

### Final Answer
The length of PQ is **15 cm**.

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