In a circle with centre 'O' PA and PB are two chords. PC is the chord that bisects the angle . APB The tangent to the circle at C is drawn meeting PA and PB extended at Q and R respectively. If QC = 3, QA = 2 and RC = 4, then length of RB equals a/b where a and b are co-prime then `a+b` is_______.

Explanation

To solve the problem, we will follow a step-by-step approach using the properties of tangents and angle bisectors in circles.

### Step 1: Understand the Geometry
We have a circle with center O, and two chords PA and PB. The chord PC bisects the angle APB. A tangent is drawn at point C, which intersects the extensions of PA and PB at points Q and R, respectively.

### Step 2: Use the Tangent-Chord Theorem
According to the tangent-chord theorem, the square of the length of the tangent segment (QC) from point Q to point C is equal to the product of the lengths of the segments of the chord (QA and QB):
\[ QC^2 = QA \cdot QB \]
Given:
- \( QC = 3 \)
- \( QA = 2 \)

We can find \( QB \):
\[ QC^2 = 3^2 = 9 \]
\[ 9 = 2 \cdot QB \implies QB = \frac{9}{2} \]

### Step 3: Apply the Angle Bisector Theorem
In triangle PQR, since PC is the angle bisector, we can use the angle bisector theorem, which states:
\[ \frac{PQ}{PR} = \frac{QC}{CR} \]
Given:
- \( QC = 3 \)
- \( CR = 4 \)
- \( PQ = QB = \frac{9}{2} \)

Let \( PR = x \). Then we can set up the equation:
\[ \frac{\frac{9}{2}}{x} = \frac{3}{4} \]

Cross-multiplying gives:
\[ 3x = 4 \cdot \frac{9}{2} \]
\[ 3x = 18 \implies x = 6 \]
So, \( PR = 6 \).

### Step 4: Use the Tangent-Chord Theorem Again
Now we will apply the tangent-chord theorem again for point R:
\[ RC^2 = RB \cdot RP \]
Given:
- \( RC = 4 \)
- \( RP = PR = 6 \)

Thus, we have:
\[ 4^2 = RB \cdot 6 \]
\[ 16 = RB \cdot 6 \implies RB = \frac{16}{6} = \frac{8}{3} \]

### Step 5: Express RB in the Form a/b
We can express \( RB \) as:
\[ RB = \frac{8}{3} \]
Here, \( a = 8 \) and \( b = 3 \), which are co-prime.

### Final Step: Calculate a + b
Now, we need to find \( a + b \):
\[ a + b = 8 + 3 = 11 \]

Thus, the final answer is:
\[ \boxed{11} \]