Find the condition that the straight line ` 3 x + 4y = k` may touch the circle ` x^(2) + y^(2) = 10 x`

Answer

To find the condition that the straight line \(3x + 4y = k\) may touch the circle \(x^2 + y^2 = 10x\), we will follow these steps:

### Step 1: Rewrite the Circle Equation
The given equation of the circle is:
\[ x^2 + y^2 = 10x \]
We can rewrite this in standard form by rearranging it:
\[ x^2 - 10x + y^2 = 0 \]
Now, we complete the square for the \(x\) terms:
\[ (x^2 - 10x + 25) + y^2 = 25 \]
This simplifies to:
\[ (x - 5)^2 + y^2 = 25 \]
This shows that the center of the circle is at \((5, 0)\) and the radius is \(5\).

### Step 2: Distance from the Center to the Line
The line \(3x + 4y = k\) can be rewritten in the standard form:
\[ 3x + 4y - k = 0 \]
To find the distance \(d\) from the center of the circle \((5, 0)\) to the line, we use the distance formula:
\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
where \(A = 3\), \(B = 4\), \(C = -k\), and \((x_1, y_1) = (5, 0)\).

Substituting these values into the formula gives:
\[ d = \frac{|3(5) + 4(0) - k|}{\sqrt{3^2 + 4^2}} = \frac{|15 - k|}{\sqrt{9 + 16}} = \frac{|15 - k|}{5} \]

### Step 3: Set the Distance Equal to the Radius
For the line to touch the circle, the distance \(d\) must be equal to the radius of the circle, which is \(5\):
\[ \frac{|15 - k|}{5} = 5 \]
Multiplying both sides by \(5\) gives:
\[ |15 - k| = 25 \]

### Step 4: Solve the Absolute Value Equation
The equation \( |15 - k| = 25 \) leads to two cases:
1. \(15 - k = 25\)
2. \(15 - k = -25\)

**Case 1:**
\[ 15 - k = 25 \implies -k = 25 - 15 \implies -k = 10 \implies k = -10 \]

**Case 2:**
\[ 15 - k = -25 \implies -k = -25 - 15 \implies -k = -40 \implies k = 40 \]

### Conclusion
The values of \(k\) for which the line \(3x + 4y = k\) touches the circle \(x^2 + y^2 = 10x\) are:
\[ k = -10 \quad \text{and} \quad k = 40 \]