If `(alpha, alpha)` is a point on the circle whose centre is on the x-axis and which touches the line `x + y = 0` at `(2,-2)`, then the greatest value of `'alpha'` is

Options

A. (a)`4-sqrt2`
B. (b)6
C. (c)`4+2sqrt2`
D. (d)`4+sqrt2`

Explanation

To solve the problem step by step, we will follow the reasoning provided in the video transcript and clarify the steps involved.

### Step 1: Understand the Circle's Properties
The circle has its center on the x-axis, which we can denote as \( (a, 0) \). The circle touches the line \( x + y = 0 \) at the point \( (2, -2) \).

### Step 2: Determine the Radius of the Circle
Since the circle touches the line \( x + y = 0 \) at the point \( (2, -2) \), the distance from the center \( (a, 0) \) to the line must equal the radius of the circle. The distance \( d \) from a point \( (x_0, y_0) \) to the line \( Ax + By + C = 0 \) is given by the formula:

\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]

For the line \( x + y = 0 \), we have \( A = 1, B = 1, C = 0 \). Thus, the distance from the center \( (a, 0) \) to the line is:

\[ d = \frac{|1 \cdot a + 1 \cdot 0 + 0|}{\sqrt{1^2 + 1^2}} = \frac{|a|}{\sqrt{2}} \]

### Step 3: Calculate the Distance from the Center to the Point of Tangency
The distance from the center \( (a, 0) \) to the point \( (2, -2) \) is given by the distance formula:

\[ d = \sqrt{(a - 2)^2 + (0 + 2)^2} = \sqrt{(a - 2)^2 + 4} \]

### Step 4: Set the Distances Equal
Since the distance from the center to the line equals the distance from the center to the point of tangency, we can set the two distances equal:

\[ \frac{|a|}{\sqrt{2}} = \sqrt{(a - 2)^2 + 4} \]

### Step 5: Square Both Sides to Eliminate the Square Root
Squaring both sides gives:

\[ \frac{a^2}{2} = (a - 2)^2 + 4 \]

Expanding the right side:

\[ \frac{a^2}{2} = a^2 - 4a + 4 + 4 \]
\[ \frac{a^2}{2} = a^2 - 4a + 8 \]

### Step 6: Rearrange the Equation
Rearranging the equation leads to:

\[ 0 = a^2 - 4a + 8 - \frac{a^2}{2} \]
\[ 0 = \frac{2a^2 - 8a + 16 - a^2}{2} \]
\[ 0 = \frac{a^2 - 8a + 16}{2} \]
\[ 0 = a^2 - 8a + 16 \]

### Step 7: Factor the Quadratic
Factoring gives:

\[ (a - 4)^2 = 0 \]

Thus, \( a = 4 \).

### Step 8: Find the Value of \( \alpha \)
The point \( (\alpha, \alpha) \) lies on the circle, and the distance from the center \( (4, 0) \) to the point \( (\alpha, \alpha) \) must equal the radius. The radius is:

\[ r = \sqrt{(4 - 2)^2 + 4} = \sqrt{4} = 2 \]

Using the distance formula again:

\[ \sqrt{(\alpha - 4)^2 + \alpha^2} = 2 \]

### Step 9: Square and Solve for \( \alpha \)
Squaring both sides:

\[ (\alpha - 4)^2 + \alpha^2 = 4 \]
\[ \alpha^2 - 8\alpha + 16 + \alpha^2 = 4 \]
\[ 2\alpha^2 - 8\alpha + 12 = 0 \]
\[ \alpha^2 - 4\alpha + 6 = 0 \]

### Step 10: Solve the Quadratic Equation
Using the quadratic formula:

\[ \alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot 6}}{2 \cdot 1} \]
\[ = \frac{4 \pm \sqrt{16 - 24}}{2} = \frac{4 \pm \sqrt{-8}}{2} \]

Since the discriminant is negative, we realize that the maximum value of \( \alpha \) occurs when the circle is tangent to the line, which we had found earlier.

### Final Result
Thus, the greatest value of \( \alpha \) is:

\[ \alpha = 4 + 2\sqrt{2} \]