If the circle `2x^(2) + 2y^(2) = 5x` touches the line `3x + 4y = k` ,then the values of `k` are

Options

A. `- 10 , (5)/(2)`
B. ` 10, - (5)/(2)`
C. `10, (5)/(2)`
D. `5,(5)/(4)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Rewrite the equation of the circle
The given equation of the circle is:
\[ 2x^2 + 2y^2 = 5x \]
We can simplify this by dividing the entire equation by 2:
\[ x^2 + y^2 = \frac{5}{2}x \]
Rearranging gives:
\[ x^2 + y^2 - \frac{5}{2}x = 0 \]

### Step 2: Identify the center and radius of the circle
The general form of a circle is:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
From our equation, we can identify:
- \( 2g = -\frac{5}{2} \) → \( g = -\frac{5}{4} \)
- \( 2f = 0 \) → \( f = 0 \)
- \( c = 0 \)

Thus, the center of the circle is:
\[ \left(-g, -f\right) = \left(\frac{5}{4}, 0\right) \]

### Step 3: Calculate the radius of the circle
The radius \( r \) can be calculated using the formula:
\[ r = \sqrt{g^2 + f^2 - c} \]
Substituting the values:
\[ r = \sqrt{\left(-\frac{5}{4}\right)^2 + 0^2 - 0} = \sqrt{\frac{25}{16}} = \frac{5}{4} \]

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

Substituting these values:
\[ d = \frac{|3 \cdot \frac{5}{4} + 4 \cdot 0 - k|}{\sqrt{3^2 + 4^2}} = \frac{| \frac{15}{4} - k |}{\sqrt{9 + 16}} = \frac{| \frac{15}{4} - k |}{5} \]

### Step 5: Set the distance equal to the radius
Since the circle touches the line, the distance \( d \) is equal to the radius \( r \):
\[ \frac{| \frac{15}{4} - k |}{5} = \frac{5}{4} \]

### Step 6: Solve for \( k \)
Multiplying both sides by 5:
\[ | \frac{15}{4} - k | = \frac{25}{4} \]

This gives us two cases to solve:
1. \( \frac{15}{4} - k = \frac{25}{4} \)
2. \( \frac{15}{4} - k = -\frac{25}{4} \)

**Case 1:**
\[ \frac{15}{4} - k = \frac{25}{4} \]
\[ -k = \frac{25}{4} - \frac{15}{4} \]
\[ -k = \frac{10}{4} \]
\[ k = -\frac{10}{4} = -\frac{5}{2} \]

**Case 2:**
\[ \frac{15}{4} - k = -\frac{25}{4} \]
\[ -k = -\frac{25}{4} - \frac{15}{4} \]
\[ -k = -\frac{40}{4} \]
\[ k = 10 \]

### Final Values of \( k \)
The values of \( k \) are:
\[ k = -\frac{5}{2} \quad \text{and} \quad k = 10 \]

### Conclusion
The values of \( k \) for which the circle touches the line are:
\[ k = -\frac{5}{2}, \quad k = 10 \]