If the line `4x + 3y +lambda=0` touches the circle `2x^(2)+2y^(2)-5x=0`, then `lambda=`………..

Answer

To find the value of \( \lambda \) such that the line \( 4x + 3y + \lambda = 0 \) touches the circle given by the equation \( 2x^2 + 2y^2 - 5x = 0 \), we can follow these steps:

### Step 1: Rewrite the equation of the circle
The given equation of the circle is:
\[ 2x^2 + 2y^2 - 5x = 0 \]
Dividing the entire equation by 2, we get:
\[ x^2 + y^2 - \frac{5}{2}x = 0 \]

### Step 2: Complete the square for the circle equation
To rewrite the equation in standard form, we complete the square for the \( x \) terms:
\[ x^2 - \frac{5}{2}x + y^2 = 0 \]
Completing the square:
\[ \left( x - \frac{5}{4} \right)^2 - \left( \frac{5}{4} \right)^2 + y^2 = 0 \]
This simplifies to:
\[ \left( x - \frac{5}{4} \right)^2 + y^2 = \left( \frac{5}{4} \right)^2 \]
Thus, the center of the circle is \( \left( \frac{5}{4}, 0 \right) \) and the radius \( r \) is \( \frac{5}{4} \).

### Step 3: Use the formula for the distance from a point to a line
The distance \( d \) from the center of the circle \( C\left( \frac{5}{4}, 0 \right) \) to the line \( 4x + 3y + \lambda = 0 \) must equal the radius for the line to be tangent to the circle. The formula for the distance from a point \( (x_0, y_0) \) to the line \( Ax + By + C = 0 \) is given by:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
Here, \( A = 4 \), \( B = 3 \), \( C = \lambda \), and \( (x_0, y_0) = \left( \frac{5}{4}, 0 \right) \).

### Step 4: Substitute values into the distance formula
Substituting the values into the distance formula:
\[ d = \frac{|4 \cdot \frac{5}{4} + 3 \cdot 0 + \lambda|}{\sqrt{4^2 + 3^2}} = \frac{|5 + \lambda|}{\sqrt{16 + 9}} = \frac{|5 + \lambda|}{5} \]

### Step 5: Set the distance equal to the radius
Setting the distance equal to the radius:
\[ \frac{|5 + \lambda|}{5} = \frac{5}{4} \]

### Step 6: Solve for \( \lambda \)
Cross-multiplying gives:
\[ |5 + \lambda| = \frac{5 \cdot 5}{4} = \frac{25}{4} \]
This leads to two cases:
1. \( 5 + \lambda = \frac{25}{4} \)
2. \( 5 + \lambda = -\frac{25}{4} \)

#### Case 1:
\[ \lambda = \frac{25}{4} - 5 = \frac{25}{4} - \frac{20}{4} = \frac{5}{4} \]

#### Case 2:
\[ \lambda = -\frac{25}{4} - 5 = -\frac{25}{4} - \frac{20}{4} = -\frac{45}{4} \]

### Conclusion
Since we are looking for the value of \( \lambda \) that allows the line to touch the circle, we take the positive value:
\[ \lambda = \frac{5}{4} \]

### Final Answer
\[ \lambda = \frac{5}{4} \]

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