At one end A of a diameter AB of a circle of radius 5cm, tangent XAY is drawn to the circle. The length of the chord CD parallel to XY and at a distance 8 cm from A is

Correct option is D

Given:

Radius of the circle = 5 cm.

Point A is the end of the diameter AB of the circle.

A tangent XAY is drawn at point A.

The distance from point A to the chord CD is 8 cm.

Chord CD is parallel to the tangent XY.

Formula Used:

The distance from the center to a chord is related to the length of the chord by the formula:

​$$d = \sqrt{r^2 - \left(\frac{l}{2}\right)^2}$$​​

where: d is the perpendicular distance from the center to the chord, r is the radius, l is the length of the chord.

Solution:

1. The distance from the center of the circle to chord CD is the radius minus the distance from A to CD:

​$$d = 5 \, \text{cm} - 8 \, \text{cm} = 3 \, \text{cm}$$​​

2. Apply the formula for the distance to the chord:

​$$d = \sqrt{r^2 - \left(\frac{l}{2}\right)^2}$$​​

Substituting $$d = 3 \, \text{cm}, r = 5 \, \text{cm}$$​:

​$$3 = \sqrt{25 - \left(\frac{l}{2}\right)^2}$$​​

3. Square both sides:

​$$9 = 25 - \left(\frac{l}{2}\right)^2$$​​

4. Solve for $$(l/2)^2$$​:

​$$\left(\frac{l}{2}\right)^2 = 16$$​​

​$$\frac{l}{2} = 4$$​​

5. Therefore, the length of chord CD is:

​$$l = 8 \, \text{cm}$$​​

Thus, the length of the chord CD is 8 cm.