If `4x+3y+k=0` touch the circle `2x^(2)+2y^(2)=5x` then k=

Options

A. `5/4` or `(-45)/4`
B. `(-5)/4` or `(-45)/4`
C. `-5/4` or `45/4`
D. `5/4` or `45/4`

Answer

To find the value of \( k \) such that the line \( 4x + 3y + k = 0 \) is tangent to the circle given by \( 2x^2 + 2y^2 = 5x \), we can follow these steps:

### Step 1: Rewrite the Circle's Equation
The equation of the circle can be rewritten in standard form. Start with:
\[ 2x^2 + 2y^2 = 5x \]
Dividing the entire equation by 2 gives:
\[ x^2 + y^2 = \frac{5}{2}x \]
Rearranging this, we have:
\[ x^2 - \frac{5}{2}x + y^2 = 0 \]

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

### Step 3: Find the Slope of the Line
From the equation of the line \( 4x + 3y + k = 0 \), we can express \( y \) in terms of \( x \):
\[ 3y = -4x - k \quad \Rightarrow \quad y = -\frac{4}{3}x - \frac{k}{3} \]
The slope of the line is \( -\frac{4}{3} \).

### Step 4: Find the Distance from the Center to the Line
The distance \( d \) from the center of the circle \( \left(\frac{5}{4}, 0\right) \) to the line \( 4x + 3y + k = 0 \) is given by the formula:
\[ d = \frac{|4\left(\frac{5}{4}\right) + 3(0) + k|}{\sqrt{4^2 + 3^2}} = \frac{|5 + k|}{5} \]

### Step 5: Set the Distance Equal to the Radius
Since the line is tangent to the circle, the distance \( d \) must equal the radius \( r \):
\[ \frac{|5 + k|}{5} = \frac{5}{4} \]
Cross-multiplying gives:
\[ |5 + k| = \frac{25}{4} \]

### Step 6: Solve for \( k \)
This absolute value equation leads to two cases:
1. \( 5 + k = \frac{25}{4} \)
2. \( 5 + k = -\frac{25}{4} \)

**For the first case:**
\[ k = \frac{25}{4} - 5 = \frac{25}{4} - \frac{20}{4} = \frac{5}{4} \]

**For the second case:**
\[ k = -\frac{25}{4} - 5 = -\frac{25}{4} - \frac{20}{4} = -\frac{45}{4} \]

### Conclusion
Thus, the values of \( k \) for which the line is tangent to the circle are:
\[ k = \frac{5}{4} \quad \text{and} \quad k = -\frac{45}{4} \]