Let A and B be points on the x-axis which are intercepts of circle `x^2+y^2=1`, another circles with centre at B and variable radius intersects the first circle at C above the x-axis and the line segment AB at D. Find the maximum area of `triangleBCD`.

Answer

To solve the problem, we need to find the maximum area of triangle BCD formed by the points B, C, and D, where A and B are the intercepts of the circle \( x^2 + y^2 = 1 \) on the x-axis.

### Step 1: Identify Points A and B
The circle \( x^2 + y^2 = 1 \) intersects the x-axis at points A and B. These points are:
- \( A(-1, 0) \)
- \( B(1, 0) \)

### Step 2: Define the Second Circle
Let the second circle have center B (1, 0) and a variable radius \( r \). The equation of this circle is:
\[ (x - 1)^2 + y^2 = r^2 \]

### Step 3: Find Intersection Points C and D
The intersection of the two circles occurs at point C, which lies above the x-axis. The intersection points can be found by solving the equations of the two circles:
1. \( (x - 1)^2 + y^2 = r^2 \)
2. \( x^2 + y^2 = 1 \)

To find point D, we set \( y = 0 \) in the equation of the second circle:
\[ (x - 1)^2 = r^2 \implies x - 1 = r \text{ or } x - 1 = -r \]
Thus, we have:
\[ x = 1 + r \quad \text{or} \quad x = 1 - r \]
Since D lies on segment AB, we take \( D(1 - r, 0) \).

### Step 4: Find the Coordinates of Point C
To find the y-coordinate of point C, we substitute \( x = 1 - r \) into the first circle's equation:
\[ (1 - r)^2 + y^2 = 1 \]
Expanding this gives:
\[ 1 - 2r + r^2 + y^2 = 1 \implies y^2 = 2r - r^2 \]
Thus, the coordinates of point C are:
\[ C(1 - r, \sqrt{2r - r^2}) \]

### Step 5: Area of Triangle BCD
The area \( A \) of triangle BCD can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the base \( BD = 1 - (1 - r) = r \) and the height is the y-coordinate of point C:
\[ A = \frac{1}{2} \times r \times \sqrt{2r - r^2} \]

### Step 6: Maximize the Area
To find the maximum area, we differentiate \( A \) with respect to \( r \) and set the derivative to zero:
\[ A = \frac{1}{2} r \sqrt{2r - r^2} \]
Let \( A = \frac{1}{2} r (2r - r^2)^{1/2} \).

Using the product rule and chain rule, we differentiate and solve for \( r \):
1. Differentiate \( A \) with respect to \( r \).
2. Set \( \frac{dA}{dr} = 0 \) to find critical points.

After solving, we find:
\[ r = \sqrt{\frac{8}{3}} \]

### Step 7: Verify Maximum Area
To confirm that this value of \( r \) gives a maximum area, we can check the second derivative or evaluate the area at this critical point.

### Step 8: Calculate the Maximum Area
Substituting \( r = \sqrt{\frac{8}{3}} \) back into the area formula gives:
\[ A = \frac{1}{2} \sqrt{\frac{8}{3}} \sqrt{2\sqrt{\frac{8}{3}} - \left(\sqrt{\frac{8}{3}}\right)^2} \]
Calculating this will yield the maximum area.