The straight-line `y= mx+ c` cuts the circle `x^(2) + y^(2) = a^(2)` in real points if `:`

Options

A. `sqrt( a^(2) (1+ m^(2))) lt c `
B. `sqrt( a^(2) (1- m^(2))) lt c `
C. `sqrt( a^(2) (1+ m^(2))) gt c `
D. `sqrt( a^(2) (1- m^(2))) gt c `

Answer

To determine the condition under which the straight line \( y = mx + c \) cuts the circle \( x^2 + y^2 = a^2 \) at real points, we need to substitute the equation of the line into the equation of the circle and analyze the resulting quadratic equation.

### Step-by-step Solution:

1. **Write the equation of the circle and the line:**
- Circle: \( x^2 + y^2 = a^2 \)
- Line: \( y = mx + c \)

2. **Substitute the line equation into the circle equation:**
- Replace \( y \) in the circle's equation:
\[ x^2 + (mx + c)^2 = a^2 \]

3. **Expand the equation:**
- Expand \( (mx + c)^2 \):
\[ x^2 + (m^2x^2 + 2mcx + c^2) = a^2 \]
- Combine like terms:
\[ (1 + m^2)x^2 + 2mcx + (c^2 - a^2) = 0 \]

4. **Identify the coefficients of the quadratic equation:**
- The quadratic equation in standard form is:
\[ Ax^2 + Bx + C = 0 \]
- Here, \( A = 1 + m^2 \), \( B = 2mc \), and \( C = c^2 - a^2 \).

5. **Determine the condition for real solutions:**
- For the quadratic equation to have real solutions, the discriminant \( D \) must be greater than or equal to zero:
\[ D = B^2 - 4AC \geq 0 \]
- Substitute the coefficients:
\[ (2mc)^2 - 4(1 + m^2)(c^2 - a^2) \geq 0 \]

6. **Simplify the discriminant:**
- Calculate \( D \):
\[ 4m^2c^2 - 4(1 + m^2)(c^2 - a^2) \geq 0 \]
- Expand:
\[ 4m^2c^2 - 4(c^2 - a^2 + m^2c^2 - m^2a^2) \geq 0 \]
- Combine like terms:
\[ 4m^2c^2 - 4c^2 + 4a^2 + 4m^2a^2 \geq 0 \]
- Factor out \( 4 \):
\[ 4(m^2c^2 - c^2 + a^2 + m^2a^2) \geq 0 \]
- This simplifies to:
\[ m^2c^2 - c^2 + a^2 + m^2a^2 \geq 0 \]

7. **Rearranging the inequality:**
- Rearranging gives:
\[ (1 + m^2)a^2 \geq c^2 \]

8. **Final condition:**
- Thus, the condition for the line to cut the circle at real points is:
\[ (1 + m^2)a^2 \geq c^2 \]