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

Options

A. `-rsqrt(1+m^(2))ltCltrsqrt(1+m^(2))`
B. `clt-rsqrt(1+m^(2))`
C. `cltrsqrt(1+m^(2))`
D. `-sqrt(1+m^(2)) lt c lt sqrt(1+m^(2))`

Answer

To determine the conditions under which the line \( y = mx + c \) intersects the circle \( x^2 + y^2 = r^2 \) at two distinct points, we can follow these steps:

### Step 1: Substitute the line equation into the circle equation
We start with the equations:
- Circle: \( x^2 + y^2 = r^2 \)
- Line: \( y = mx + c \)

Substituting the line equation into the circle equation gives:
\[ x^2 + (mx + c)^2 = r^2 \]

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

### Step 3: Identify the quadratic form
We can identify this as a quadratic equation in \( x \):
\[ Ax^2 + Bx + C = 0 \]
where:
- \( A = 1 + m^2 \)
- \( B = 2mc \)
- \( C = c^2 - r^2 \)

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

### Step 5: Simplify the discriminant
Calculating the discriminant:
\[ 4m^2c^2 - 4(1 + m^2)(c^2 - r^2) > 0 \]
This simplifies to:
\[ 4m^2c^2 - 4(c^2 - r^2) - 4m^2c^2 > 0 \]
Thus,
\[ 4r^2 - 4c^2 > 0 \]
or
\[ r^2 > c^2 \]

### Step 6: Rearranging the inequality
Taking square roots gives:
\[ |c| < r \]
This means:
\[ -r < c < r \]

### Step 7: Find the final condition
To find the specific conditions for \( c \), we need to consider the slope \( m \):
\[ -r\sqrt{1 + m^2} < c < r\sqrt{1 + m^2} \]

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