The radius of a circle, having minimum area, which touches the curve `y=4-x^2` and the lines `y=|x|` is :

Options

A. (a) `4(sqrt(2)+1)`
B. (b) `2(sqrt(2)+1)`
C. (c) `2(sqrt(2)-1)`
D. (d) `4(sqrt(2)-1)`

Explanation

To solve the problem of finding the radius of a circle with minimum area that touches the curve \( y = 4 - x^2 \) and the lines \( y = |x| \), we can follow these steps:

### Step 1: Understand the curves and lines
The curve \( y = 4 - x^2 \) is a downward-opening parabola with its vertex at (0, 4). The lines \( y = |x| \) consist of two lines: \( y = x \) for \( x \geq 0 \) and \( y = -x \) for \( x < 0 \).

### Step 2: Identify the center of the circle
Let the center of the circle be at the point \( (0, 4 - r) \), where \( r \) is the radius of the circle. The circle will touch the parabola and the lines \( y = x \) and \( y = -x \).

### Step 3: Find the distance from the center to the lines
The distance from the center of the circle \( (0, 4 - r) \) to the line \( y = x \) can be calculated using the formula for the distance from a point to a line. The line \( y = x \) can be rewritten as \( x - y = 0 \). The distance \( d \) from the point \( (0, 4 - r) \) to the line \( x - y = 0 \) is given by:

\[ d = \frac{|0 - (4 - r)|}{\sqrt{1^2 + (-1)^2}} = \frac{|4 - r|}{\sqrt{2}} \]

### Step 4: Set the distance equal to the radius
Since the circle touches the line \( y = x \), this distance must equal the radius \( r \):

\[ \frac{|4 - r|}{\sqrt{2}} = r \]

### Step 5: Solve for \( r \)
We can solve the equation \( |4 - r| = r\sqrt{2} \). This gives us two cases to consider:

**Case 1:** \( 4 - r = r\sqrt{2} \)

\[ 4 = r + r\sqrt{2} \implies 4 = r(1 + \sqrt{2}) \implies r = \frac{4}{1 + \sqrt{2}} \]

**Case 2:** \( 4 - r = -r\sqrt{2} \)

\[ 4 = r - r\sqrt{2} \implies 4 = r(1 - \sqrt{2}) \]
Since \( 1 - \sqrt{2} < 0 \), this case will give a negative radius, which is not possible.

### Step 6: Rationalize the radius
Now we need to rationalize \( r = \frac{4}{1 + \sqrt{2}} \):

\[ r = \frac{4(1 - \sqrt{2})}{(1 + \sqrt{2})(1 - \sqrt{2})} = \frac{4(1 - \sqrt{2})}{1 - 2} = -4(1 - \sqrt{2}) = 4\sqrt{2} - 4 \]

### Final Answer
Thus, the radius of the circle having minimum area that touches the curve \( y = 4 - x^2 \) and the lines \( y = |x| \) is:

\[ r = 4\sqrt{2} - 4 \]