Two circles cut each other at points P and Q. Centres of two circles are respectively A and B and PA is perpendicular to PB. If AB intersects segment PQ at R ratio of the two circles are respectively `16:9` then `AR:BR` is

Options

A. `4:3`
B. `16:9`
C. `256:81`
D. `2: sqrt(3)`

Answer

To solve the problem step by step, we will use the properties of triangles and the concept of the ratio of areas.

### Step-by-Step Solution:

1. **Understanding the Configuration**:
We have two circles that intersect at points P and Q. Let A and B be the centers of the two circles. The line segment PA is perpendicular to PB.

2. **Identifying the Ratios**:
We are given that the ratio of the circles is 16:9. This means that the radius of circle A (R1) and the radius of circle B (R2) can be expressed as:
\[ \frac{R1}{R2} = \frac{4}{3} \]
(since \( \sqrt{16} = 4 \) and \( \sqrt{9} = 3 \)).

3. **Using Triangle Areas**:
We will use the areas of triangles ARP and BRP. The area of triangle ARP can be expressed as:
\[ \text{Area}_{ARP} = \frac{1}{2} \cdot AR \cdot AP \]
and the area of triangle BRP as:
\[ \text{Area}_{BRP} = \frac{1}{2} \cdot BR \cdot BP \]

4. **Setting Up the Ratio of Areas**:
The ratio of the areas of triangles ARP and BRP can be set up as:
\[ \frac{\text{Area}_{ARP}}{\text{Area}_{BRP}} = \frac{AR \cdot AP}{BR \cdot BP} \]

5. **Using the Power of a Point Theorem**:
According to the Power of a Point theorem, we have:
\[ \frac{AP}{PB} = \left(\frac{R1}{R2}\right)^2 \]
Substituting the ratio of the radii:
\[ \frac{AP}{PB} = \frac{16}{9} \]

6. **Substituting into the Area Ratio**:
Now we can substitute this into our area ratio:
\[ \frac{AR \cdot AP}{BR \cdot BP} = \frac{AR}{BR} \cdot \frac{AP}{BP} \]
From the Power of a Point theorem, we have:
\[ \frac{AP}{BP} = \frac{16}{9} \]

7. **Final Ratio Calculation**:
Thus, we can express the ratio of AR to BR:
\[ \frac{AR}{BR} = \frac{AP}{PB} = \frac{16}{9} \]
Therefore, we have:
\[ AR:BR = 256:81 \]

### Conclusion:
The ratio \( AR:BR \) is \( 256:81 \).