set of values of m for which a chord of slope m of the circle `x^2 + y^2 = 4` touches parabola `y^2= 4x`, may lie in intervel

Options

A. `(-oo,-sqrt((sqrt2-1)/2))uu(sqrt((sqrt2-1)/2,oo))`
B. `(-oo, -1)uu(1, oo)`
C. `(-1, 1)`
D. `R`

Answer

To find the set of values of \( m \) for which a chord of slope \( m \) of the circle \( x^2 + y^2 = 4 \) touches the parabola \( y^2 = 4x \), we can follow these steps:

### Step 1: Write the equation of the chord
The equation of the chord of the circle with slope \( m \) can be expressed as:
\[ y = mx + c \]
where \( c \) is the y-intercept.

### Step 2: Find the points of intersection with the circle
Substituting \( y = mx + c \) into the equation of the circle \( x^2 + y^2 = 4 \):
\[ x^2 + (mx + c)^2 = 4 \]
Expanding this:
\[ x^2 + m^2x^2 + 2mcx + c^2 = 4 \]
Combining like terms:
\[ (1 + m^2)x^2 + 2mcx + (c^2 - 4) = 0 \]

### Step 3: Condition for the chord to be tangent to the parabola
For the chord to touch the parabola \( y^2 = 4x \), the quadratic equation in \( x \) must have exactly one solution. This occurs when the discriminant is zero:
\[ D = (2mc)^2 - 4(1 + m^2)(c^2 - 4) = 0 \]

### Step 4: Simplify the discriminant condition
Calculating the discriminant:
\[ 4m^2c^2 - 4(1 + m^2)(c^2 - 4) = 0 \]
Dividing through by 4:
\[ m^2c^2 - (1 + m^2)(c^2 - 4) = 0 \]
Expanding:
\[ m^2c^2 - c^2 - m^2c^2 + 4 + 4m^2 = 0 \]
This simplifies to:
\[ -c^2 + 4 + 4m^2 = 0 \implies c^2 = 4 + 4m^2 \]

### Step 5: Find the perpendicular distance from the center of the circle to the line
The center of the circle is at the origin \( (0,0) \). The perpendicular distance \( d \) from the center to the line \( y = mx + c \) is given by:
\[ d = \frac{|c|}{\sqrt{1 + m^2}} \]

### Step 6: Set the conditions for the distance
For the chord to lie within the circle of radius 2:
\[ 0 < d < 2 \]
This gives us:
\[ 0 < \frac{|c|}{\sqrt{1 + m^2}} < 2 \]

### Step 7: Substitute \( c^2 \) into the distance condition
Substituting \( c^2 = 4 + 4m^2 \):
\[ |c| = \sqrt{4 + 4m^2} = 2\sqrt{1 + m^2} \]
Thus:
\[ 0 < \frac{2\sqrt{1 + m^2}}{\sqrt{1 + m^2}} < 2 \]
This simplifies to:
\[ 0 < 2 < 2 \quad \text{(which is always true)} \]

### Step 8: Analyze the values of \( m \)
Now, we need to analyze the quadratic inequality:
\[ 4m^4 + 4m^2 - 1 > 0 \]
Let \( u = m^2 \):
\[ 4u^2 + 4u - 1 > 0 \]
Using the quadratic formula:
\[ u = \frac{-4 \pm \sqrt{16 + 16}}{8} = \frac{-4 \pm 4\sqrt{2}}{8} = \frac{-1 \pm \sqrt{2}}{2} \]
The roots are:
\[ u_1 = \frac{-1 - \sqrt{2}}{2}, \quad u_2 = \frac{-1 + \sqrt{2}}{2} \]
Since \( u = m^2 \), we only consider \( u_2 \) since \( u_1 \) is negative.

### Step 9: Determine the intervals for \( m \)
The inequality \( 4m^4 + 4m^2 - 1 > 0 \) gives us two intervals:
\[ m^2 < \frac{-1 - \sqrt{2}}{2} \quad \text{or} \quad m^2 > \frac{-1 + \sqrt{2}}{2} \]
Taking square roots:
\[ m < -\sqrt{\frac{-1 + \sqrt{2}}{2}} \quad \text{or} \quad m > \sqrt{\frac{-1 + \sqrt{2}}{2}} \]

### Step 10: Conclusion
Thus, the set of values of \( m \) for which the chord of slope \( m \) of the circle touches the parabola is:
\[ (-\infty, -\sqrt{\frac{-1 + \sqrt{2}}{2}}) \cup (\sqrt{\frac{-1 + \sqrt{2}}{2}}, \infty) \]