If `sintheta ` is the acute angle between the curves ` x^(2)+y^(2)=4x " and " x^(2)+y^(2)=8 " at " (2,2),` then `theta=`

Options

A. 1
B. 0
C. `1//sqrt(2)`
D. `sqrt(3)//2`

Explanation

To find the acute angle \( \theta \) between the curves \( x^2 + y^2 = 4x \) and \( x^2 + y^2 = 8 \) at the point \( (2, 2) \), we will follow these steps:

### Step 1: Differentiate the first curve
The first curve is given by:
\[ x^2 + y^2 = 4x \]
Differentiating both sides with respect to \( x \):
\[ 2x + 2y \frac{dy}{dx} = 4 \]
Rearranging gives:
\[ 2y \frac{dy}{dx} = 4 - 2x \]
Thus,
\[ \frac{dy}{dx} = \frac{4 - 2x}{2y} \]

### Step 2: Substitute the point (2, 2) into the derivative
Substituting \( x = 2 \) and \( y = 2 \):
\[ \frac{dy}{dx} = \frac{4 - 2(2)}{2(2)} = \frac{4 - 4}{4} = \frac{0}{4} = 0 \]
So, the slope \( M_1 \) of the first curve at the point \( (2, 2) \) is \( 0 \).

### Step 3: Differentiate the second curve
The second curve is given by:
\[ x^2 + y^2 = 8 \]
Differentiating both sides with respect to \( x \):
\[ 2x + 2y \frac{dy}{dx} = 0 \]
Rearranging gives:
\[ 2y \frac{dy}{dx} = -2x \]
Thus,
\[ \frac{dy}{dx} = -\frac{2x}{2y} = -\frac{x}{y} \]

### Step 4: Substitute the point (2, 2) into the derivative
Substituting \( x = 2 \) and \( y = 2 \):
\[ \frac{dy}{dx} = -\frac{2}{2} = -1 \]
So, the slope \( M_2 \) of the second curve at the point \( (2, 2) \) is \( -1 \).

### Step 5: Calculate the tangent of the angle between the curves
Using the formula for the tangent of the angle \( \theta \) between two curves:
\[ \tan \theta = \frac{M_1 - M_2}{1 + M_1 M_2} \]
Substituting \( M_1 = 0 \) and \( M_2 = -1 \):
\[ \tan \theta = \frac{0 - (-1)}{1 + 0 \cdot (-1)} = \frac{1}{1} = 1 \]

### Step 6: Find the angle \( \theta \)
Since \( \tan \theta = 1 \), we have:
\[ \theta = \frac{\pi}{4} \]

### Step 7: Find \( \sin \theta \)
The problem asks for \( \sin \theta \):
\[ \sin \theta = \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]

### Final Answer
Thus, the value of \( \sin \theta \) is:
\[ \sin \theta = \frac{1}{\sqrt{2}} \]

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