`AB = 2r `is the diameter of a circle. If a chord CD Interests AB at right angle at a point P in the ratio `1:2`, then CD is equal to

Options

A. `(2sqrt(2))/3 r`
B. `(4r)/3`
C. `(4sqrt(2)r)/(3)`
D. `(4sqrt(3)r)/(3)`

Answer

To solve the problem step by step, we will follow the reasoning provided in the video transcript.

### Step-by-Step Solution:

1. **Understanding the Circle and Diameter**:
- Given that \( AB = 2r \) is the diameter of the circle, we can identify the center \( O \) of the circle. The radius \( OA = OB = r \).

2. **Identifying the Point of Intersection**:
- The chord \( CD \) intersects the diameter \( AB \) at point \( P \) at a right angle. The ratio in which point \( P \) divides the diameter \( AB \) is \( 1:2 \).

3. **Calculating the Lengths of AP and PB**:
- Since \( AB \) is divided in the ratio \( 1:2 \), we can denote the lengths:
- Let \( AP = x \) and \( PB = 2x \).
- Therefore, \( AB = AP + PB = x + 2x = 3x \).
- Given \( AB = 2r \), we have:
\[ 3x = 2r \implies x = \frac{2r}{3} \]
- Thus, \( AP = \frac{2r}{3} \) and \( PB = \frac{4r}{3} \).

4. **Finding the Length of OP**:
- The length \( OP \) can be calculated as:
\[ OP = OA - AP = r - \frac{2r}{3} = \frac{r}{3} \]

5. **Using Pythagorean Theorem**:
- In triangle \( OCP \) (where \( C \) is a point on the chord \( CD \)):
- \( OC = r \) (the radius),
- \( OP = \frac{r}{3} \).
- By the Pythagorean theorem:
\[ OC^2 = OP^2 + CP^2 \]
\[ r^2 = \left(\frac{r}{3}\right)^2 + CP^2 \]
\[ r^2 = \frac{r^2}{9} + CP^2 \]
- Rearranging gives:
\[ CP^2 = r^2 - \frac{r^2}{9} = \frac{9r^2}{9} - \frac{r^2}{9} = \frac{8r^2}{9} \]
- Taking the square root:
\[ CP = \frac{r\sqrt{8}}{3} = \frac{2r\sqrt{2}}{3} \]

6. **Finding Length of Chord CD**:
- Since \( CP = PD \) (because \( CD \) is perpendicular to \( AB \)):
\[ CD = CP + PD = 2 \times CP = 2 \times \frac{2r\sqrt{2}}{3} = \frac{4r\sqrt{2}}{3} \]

### Final Answer:
Thus, the length of the chord \( CD \) is:
\[ CD = \frac{4r\sqrt{2}}{3} \]