The line `y=mx+c` intersects the circle `x^(2)+y^(2)=a^(2)` in two distinct real points if

Options

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

Answer

To determine the conditions under which the line \( y = mx + c \) intersects the circle \( x^2 + y^2 = a^2 \) at two distinct 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:
\[ y = mx + c \]
Now, substitute \( y \) into the circle equation:
\[ x^2 + (mx + c)^2 = a^2 \]

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

### Step 3: Identify the coefficients
This is a quadratic equation in the standard form \( Ax^2 + Bx + C = 0 \), where:
- \( A = 1 + m^2 \)
- \( B = 2mc \)
- \( C = c^2 - a^2 \)

### Step 4: Apply the discriminant condition
For the quadratic equation to have two distinct real roots, the discriminant \( D \) must be greater than zero:
\[ D = B^2 - 4AC > 0 \]
Substituting the values of \( A \), \( B \), and \( C \):
\[ (2mc)^2 - 4(1 + m^2)(c^2 - a^2) > 0 \]

### Step 5: Simplify the discriminant
Calculating the discriminant:
\[ 4m^2c^2 - 4(1 + m^2)(c^2 - a^2) > 0 \]
Expanding the second term:
\[ 4m^2c^2 - 4(c^2 - a^2 + m^2c^2) > 0 \]
This simplifies to:
\[ 4m^2c^2 - 4c^2 + 4a^2 + 4m^2c^2 > 0 \]
Combining like terms:
\[ 8m^2c^2 - 4c^2 + 4a^2 > 0 \]
Factoring out 4:
\[ 4(2m^2c^2 - c^2 + a^2) > 0 \]
This leads to:
\[ 2m^2c^2 - c^2 + a^2 > 0 \]

### Step 6: Rearrange the inequality
Rearranging gives:
\[ c^2(2m^2 - 1) < a^2 \]

### Step 7: Determine the condition for \( c \)
To find the condition on \( c \):
\[ c^2 < \frac{a^2}{2m^2 - 1} \quad \text{(if \( 2m^2 - 1 > 0 \))} \]
This implies:
\[ c < a \sqrt{\frac{1}{2m^2 - 1}} \quad \text{(if \( 2m^2 - 1 > 0 \))} \]
If \( 2m^2 - 1 < 0 \), then \( c \) can take any value.

### Final Condition
Thus, the line \( y = mx + c \) intersects the circle \( x^2 + y^2 = a^2 \) at two distinct real points if:
\[ c < a \sqrt{1 + m^2} \]