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 conditions under which the straight line \( y = mx + c \) intersects the circle \( x^2 + y^2 = a^2 \) in real points, we can follow these steps:

### Step 1: Substitute the line equation into the circle equation
We start with the equation of the line:
\[ y = mx + c \]
Substituting this into the circle's equation \( x^2 + y^2 = a^2 \):
\[ x^2 + (mx + c)^2 = a^2 \]

### Step 2: Expand the equation
Expanding the left-hand side:
\[ x^2 + (m^2x^2 + 2mcx + c^2) = a^2 \]
This simplifies to:
\[ (1 + m^2)x^2 + 2mcx + (c^2 - a^2) = 0 \]

### Step 3: Identify the quadratic equation
The equation we have is a quadratic in \( x \):
\[ (1 + m^2)x^2 + 2mcx + (c^2 - a^2) = 0 \]

### Step 4: Calculate the discriminant
For the line to intersect the circle in real points, the discriminant of this quadratic equation must be non-negative. The discriminant \( D \) is given by:
\[ D = b^2 - 4ac \]
In our case:
- \( a = 1 + m^2 \)
- \( b = 2mc \)
- \( c = c^2 - a^2 \)

Thus, the discriminant becomes:
\[ D = (2mc)^2 - 4(1 + m^2)(c^2 - a^2) \]

### Step 5: Set the discriminant greater than or equal to zero
We require:
\[ D \geq 0 \]
This gives us:
\[ 4m^2c^2 - 4(1 + m^2)(c^2 - a^2) \geq 0 \]

### Step 6: Simplify the inequality
Expanding this inequality:
\[ 4m^2c^2 - 4(c^2 - a^2 + m^2c^2) \geq 0 \]
This simplifies to:
\[ 4m^2c^2 - 4c^2 + 4a^2 \geq 0 \]
Factoring out 4:
\[ 4(m^2 - 1)c^2 + 4a^2 \geq 0 \]

### Step 7: Rearranging the terms
Dividing through by 4:
\[ (m^2 - 1)c^2 + a^2 \geq 0 \]

### Step 8: Conclusion
For this inequality to hold true, we can conclude:
\[ c^2 \leq a^2(1 + m^2) \]
Thus, the condition for the line \( y = mx + c \) to intersect the circle \( x^2 + y^2 = a^2 \) in real points is:
\[ |c| \leq a\sqrt{1 + m^2} \]