If `4a^2+c^2=b^2-4a c ,` then the variable line `a x+b y+c=0` always passes through two fixed points. The coordinates of the fixed points can be (a)`(-2,-1)` (b) `(2,-1)` (c)`(-2,1)` (d) `(2,1)`

Options

A. `(-2,-1)`
B. `(2,-1)`
C. `(-2,1)`
D. `(2,1)`

Answer

To solve the problem, we start with the given equation:

\[ 4a^2 + c^2 = b^2 - 4ac. \]

### Step 1: Rearranging the Equation
We can rearrange the equation to isolate terms involving \( b^2 \):

\[ b^2 = 4a^2 + c^2 + 4ac. \]

### Step 2: Completing the Square
Next, we can rewrite the right-hand side to complete the square for the terms involving \( a \) and \( c \):

\[ b^2 = 4a^2 + 4ac + c^2 = (2a + c)^2. \]

### Step 3: Setting Up the Line Equation
Now, we consider the variable line given by:

\[ ax + by + c = 0. \]

### Step 4: Finding Fixed Points
To find the fixed points through which this line always passes, we can analyze the condition derived from the equation \( b^2 = (2a + c)^2 \).

From the equation \( ax + by + c = 0 \), we can express \( y \) in terms of \( x \):

\[ y = -\frac{a}{b}x - \frac{c}{b}. \]

### Step 5: Substituting Values
To find the fixed points, we can substitute specific values for \( x \) and \( y \) to check which of the given options satisfy the line equation.

Let’s check the options provided:

1. For the point \( (2, -1) \):
\[ a(2) + b(-1) + c = 0 \implies 2a - b + c = 0. \]
This can be satisfied based on the values of \( a, b, c \).

2. For the point \( (2, 1) \):
\[ a(2) + b(1) + c = 0 \implies 2a + b + c = 0. \]
This can also be satisfied.

3. For the point \( (-2, -1) \):
\[ a(-2) + b(-1) + c = 0 \implies -2a - b + c = 0. \]
This can be satisfied as well.

4. For the point \( (-2, 1) \):
\[ a(-2) + b(1) + c = 0 \implies -2a + b + c = 0. \]
This can also be satisfied.

### Conclusion
Thus, the fixed points through which the line always passes can be:

- \( (2, -1) \)
- \( (2, 1) \)

The correct options are:
(b) \( (2, -1) \) and (d) \( (2, 1) \).