If the lines joining the origin to the points of intersection of the line `y=mx+2` and the curve `x^(2)+y^(2)=1` are at right-angles, then

Options

A. `m^(2)=1`
B. `m^(2)=2`
C. `m^(2)=7`
D. `2m^(2)=1`

Explanation

To solve the problem, we need to find the value of \( m \) such that the lines joining the origin to the points of intersection of the line \( y = mx + 2 \) and the curve \( x^2 + y^2 = 1 \) are at right angles.

### Step 1: Substitute the line equation into the curve equation
We start with the line equation:
\[ y = mx + 2 \]
We substitute this into the curve equation:
\[ x^2 + y^2 = 1 \]
This gives us:
\[ x^2 + (mx + 2)^2 = 1 \]

### Step 2: Expand the equation
Expanding \( (mx + 2)^2 \):
\[ x^2 + (m^2x^2 + 4mx + 4) = 1 \]
Combining terms:
\[ (1 + m^2)x^2 + 4mx + 4 - 1 = 0 \]
This simplifies to:
\[ (1 + m^2)x^2 + 4mx + 3 = 0 \]

### Step 3: Identify coefficients
In the quadratic equation \( ax^2 + bx + c = 0 \), we have:
- \( a = 1 + m^2 \)
- \( b = 4m \)
- \( c = 3 \)

### Step 4: Condition for lines to be at right angles
For the lines joining the origin to the points of intersection to be at right angles, the condition is:
\[ b^2 - 4ac = 0 \]
This means the discriminant must be zero for the lines to be perpendicular.

### Step 5: Set up the equation
Substituting the values of \( a \), \( b \), and \( c \):
\[ (4m)^2 - 4(1 + m^2)(3) = 0 \]
This simplifies to:
\[ 16m^2 - 12 - 12m^2 = 0 \]
Combining like terms:
\[ 4m^2 - 12 = 0 \]

### Step 6: Solve for \( m^2 \)
Rearranging gives:
\[ 4m^2 = 12 \]
Dividing by 4:
\[ m^2 = 3 \]

### Step 7: Conclusion
Thus, the required condition is:
\[ m^2 = 3 \]