A line is tangent to a circle if the length of perpendicular from the centre of the circle to the line is equal to the radius of the circle. If `4l^2 - 5m^2 + 6l + 1 = 0`, then the line `lx + my + 1=0` touches a fixed circle whose centre. (A) Lies on x-axis (B) lies on yl-axis (C) is origin (D) none of these

Options

A. The centre of the circle is equal to `sqrt( 5)`
B. The radius of the circle is equal to `sqrt( 5)`
C. The circle passes through origin
D. None of these

Answer

To solve the problem, we need to analyze the given equation and the conditions for a line to be tangent to a circle.

### Step-by-Step Solution:

1. **Identify the given equations:**
We have the equation \( 4l^2 - 5m^2 + 6l + 1 = 0 \) and the line equation \( lx + my + 1 = 0 \).

2. **Understand the condition for tangency:**
A line \( Ax + By + C = 0 \) is tangent to a circle with center \( (h, k) \) and radius \( r \) if the perpendicular distance from the center of the circle to the line is equal to the radius. The distance \( D \) from the point \( (h, k) \) to the line is given by:
\[ D = \frac{|Ah + Bk + C|}{\sqrt{A^2 + B^2}} \]
In our case, \( A = l \), \( B = m \), and \( C = 1 \).

3. **Set up the distance formula:**
Let the center of the circle be \( (α, β) \) and the radius be \( R \). The distance from the center to the line is:
\[ D = \frac{|lα + mβ + 1|}{\sqrt{l^2 + m^2}} \]
For the line to be tangent to the circle, we need:
\[ D = R \]

4. **Square both sides:**
Squaring both sides gives us:
\[ (lα + mβ + 1)^2 = R^2(l^2 + m^2) \]

5. **Expand both sides:**
Expanding the left side:
\[ l^2α^2 + m^2β^2 + 1 + 2lα + 2mβ + 2lmαβ = R^2(l^2 + m^2) \]

6. **Rearranging terms:**
Rearranging gives:
\[ l^2α^2 - R^2l^2 + m^2β^2 - R^2m^2 + 2lα + 2mβ + 2lmαβ + 1 = 0 \]

7. **Comparing with the given equation:**
We compare this with the equation \( 4l^2 - 5m^2 + 6l + 1 = 0 \). This indicates that the coefficients of \( l^2 \), \( m^2 \), \( l \), and the constant term must match.

8. **Finding conditions for \( α \) and \( β \):**
From the comparison, we can derive:
- The coefficient of \( l \) gives \( 2α = 6 \) leading to \( α = 3 \).
- The coefficient of \( m \) gives \( 2β = 0 \) leading to \( β = 0 \).
- The terms involving \( l^2 \) and \( m^2 \) will help us determine if the center lies on the x-axis or y-axis.

9. **Conclusion:**
Since \( β = 0 \), this means the center of the circle lies on the x-axis.

### Final Answer:
The center of the fixed circle lies on the x-axis.