To solve the problem, we need to determine the values of \( \alpha \) for which the line \( |y| = x - \alpha \) does not intersect the circle given by the equation \( x^2 + y^2 - 10x + 21 = 0 \).
### Step 1: Rewrite the circle's equation in standard form
The given equation of the circle is:
\[
x^2 + y^2 - 10x + 21 = 0
\]
We can complete the square for the \( x \) terms:
\[
(x^2 - 10x) + y^2 + 21 = 0
\]
\[
(x - 5)^2 - 25 + y^2 + 21 = 0
\]
\[
(x - 5)^2 + y^2 - 4 = 0
\]
Thus, the equation of the circle can be rewritten as:
\[
(x - 5)^2 + y^2 = 4
\]
This shows that the center of the circle is \( (5, 0) \) and the radius \( r \) is \( 2 \).
### Step 2: Find the perpendicular distance from the center to the line
The line given is \( |y| = x - \alpha \), which can be expressed as two lines:
1. \( y = x - \alpha \)
2. \( y = - (x - \alpha) = -x + \alpha \)
We will calculate the perpendicular distance from the center of the circle \( (5, 0) \) to the line \( y = x - \alpha \).
The general form of a line is \( Ax + By + C = 0 \). For the line \( y = x - \alpha \), we can rewrite it as:
\[
x - y - \alpha = 0
\]
Here, \( A = 1 \), \( B = -1 \), and \( C = -\alpha \).
The formula for the perpendicular distance \( d \) from a point \( (x_1, y_1) \) to the line \( Ax + By + C = 0 \) is given by:
\[
d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}
\]
Substituting \( (x_1, y_1) = (5, 0) \):
\[
d = \frac{|1 \cdot 5 + (-1) \cdot 0 - \alpha|}{\sqrt{1^2 + (-1)^2}} = \frac{|5 - \alpha|}{\sqrt{2}}
\]
### Step 3: Set up the inequality for non-intersection
For the line not to intersect the circle, the perpendicular distance must be greater than the radius:
\[
\frac{|5 - \alpha|}{\sqrt{2}} > 2
\]
Multiplying both sides by \( \sqrt{2} \):
\[
|5 - \alpha| > 2\sqrt{2}
\]
### Step 4: Solve the absolute value inequality
This absolute value inequality can be split into two cases:
1. \( 5 - \alpha > 2\sqrt{2} \)
2. \( 5 - \alpha < -2\sqrt{2} \)
**Case 1:**
\[
5 - \alpha > 2\sqrt{2} \implies \alpha < 5 - 2\sqrt{2}
\]
**Case 2:**
\[
5 - \alpha < -2\sqrt{2} \implies \alpha > 5 + 2\sqrt{2}
\]
### Step 5: Combine the results
Thus, the values of \( \alpha \) that satisfy the condition are:
\[
\alpha < 5 - 2\sqrt{2} \quad \text{or} \quad \alpha > 5 + 2\sqrt{2}
\]
### Conclusion
Since \( \alpha > 0 \), we can conclude that \( \alpha \) belongs to the intervals:
\[
(0, 5 - 2\sqrt{2}) \cup (5 + 2\sqrt{2}, \infty)
\]
### Final Answer
Thus, the correct option is:
**Option 1: \( (0, 5 - 2\sqrt{2}) \cup (5 + 2\sqrt{2}, \infty) \)**
