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((sqrt(2)-1))/(2))) cup (sqrt((sqrt(2)-1))/(2),oo)`
B. `(-oo,-1) cup (1,oo)`
C. `(-1,1)`
D. `R`

Answer

To solve the problem of finding 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: Understand the equations
The given circle is \( x^2 + y^2 = 4 \), which has a radius of 2. The parabola is given by \( y^2 = 4x \).

### Step 2: Find the equation of the chord
A chord of the circle with slope \( m \) can be expressed in the form:
\[ y = mx + c \]
To find the value of \( c \), we can use the condition that this line intersects the circle. The equation of the circle can be rewritten as:
\[ y^2 = 4 - x^2 \]
Substituting \( y = mx + c \) into the circle's equation gives:
\[ (mx + c)^2 + x^2 = 4 \]

### Step 3: Expand and rearrange
Expanding the equation:
\[ m^2x^2 + 2mcx + c^2 + x^2 = 4 \]
Combine like terms:
\[ (m^2 + 1)x^2 + 2mcx + (c^2 - 4) = 0 \]

### Step 4: Condition for tangency
For the line to be a tangent to the parabola, the discriminant of this quadratic equation must be zero:
\[ (2mc)^2 - 4(m^2 + 1)(c^2 - 4) = 0 \]

### Step 5: Simplify the discriminant
Expanding the discriminant:
\[ 4m^2c^2 - 4(m^2 + 1)(c^2 - 4) = 0 \]
\[ 4m^2c^2 - 4(m^2c^2 + c^2 - 4m^2 - 4) = 0 \]
\[ 4m^2c^2 - 4m^2c^2 - 4c^2 + 16m^2 + 16 = 0 \]
This simplifies to:
\[ -4c^2 + 16m^2 + 16 = 0 \]
Dividing through by -4 gives:
\[ c^2 = 4m^2 + 4 \]

### Step 6: Substitute back to find \( c \)
Taking the square root:
\[ c = \pm \sqrt{4m^2 + 4} = \pm 2\sqrt{m^2 + 1} \]

### Step 7: Find distance from center to line
The distance \( d \) from the center of the circle (0,0) to the line \( y = mx + c \) is given by:
\[ d = \frac{|c|}{\sqrt{1 + m^2}} \]
For the chord to lie within the circle, this distance must be less than or equal to the radius (2):
\[ \frac{|c|}{\sqrt{1 + m^2}} \leq 2 \]
Substituting \( c \):
\[ \frac{2\sqrt{m^2 + 1}}{\sqrt{1 + m^2}} \leq 2 \]

### Step 8: Square both sides
Squaring both sides gives:
\[ 4(m^2 + 1) \leq 4(1 + m^2) \]
This simplifies to:
\[ 4m^2 + 4 \leq 4 + 4m^2 \]
This is always true, indicating that the condition is satisfied for all \( m \).

### Step 9: Final solution
Thus, the set of values of \( m \) for which the chord of slope \( m \) of the circle touches the parabola is:
\[ m \in (-\infty, -\sqrt{2}) \cup (\sqrt{2}, \infty) \]