If line ` 4x + 3y + k = 0 " touches circle " 2x^(2) + 2y^(2) = 5x ` , then k =

Options

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

Answer

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

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

### Step 2: Identify the center and radius of the circle
The general form of a circle is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
To convert our equation into this form, we complete the square for the \( x \) term:
\[ x^2 - \frac{5}{2}x = \left(x - \frac{5}{4}\right)^2 - \left(\frac{5}{4}\right)^2 \]
Thus, we can rewrite the equation as:
\[ \left(x - \frac{5}{4}\right)^2 + y^2 = \left(\frac{5}{4}\right)^2 \]
From this, we identify:
- Center \( C\left(\frac{5}{4}, 0\right) \)
- Radius \( r = \frac{5}{4} \)

### Step 3: Find the perpendicular distance from the center to the line
The formula for the perpendicular distance \( d \) from a point \( (x_1, y_1) \) to the line \( Ax + By + C = 0 \) is:
\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
Here, \( A = 4 \), \( B = 3 \), \( C = k \), and the point is \( \left(\frac{5}{4}, 0\right) \).

Calculating the distance:
\[ d = \frac{|4 \cdot \frac{5}{4} + 3 \cdot 0 + k|}{\sqrt{4^2 + 3^2}} = \frac{|5 + k|}{\sqrt{16 + 9}} = \frac{|5 + k|}{5} \]

### Step 4: Set the distance equal to the radius
Since the line touches the circle, the distance \( d \) must equal the radius \( r \):
\[ \frac{|5 + k|}{5} = \frac{5}{4} \]

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

**Case 1:**
\[ 5 + k = \frac{25}{4} \implies k = \frac{25}{4} - 5 = \frac{25}{4} - \frac{20}{4} = \frac{5}{4} \]

**Case 2:**
\[ 5 + k = -\frac{25}{4} \implies k = -\frac{25}{4} - 5 = -\frac{25}{4} - \frac{20}{4} = -\frac{45}{4} \]

### Step 6: Conclusion
The possible values for \( k \) are \( \frac{5}{4} \) and \( -\frac{45}{4} \). Since the problem states that the line touches the circle, we need to check which of these is valid.

The valid value of \( k \) that corresponds to the tangent condition is:
\[ \boxed{-\frac{45}{4}} \]