1. Set up the geometry

Let the required points on the line $y=2$ be $A(x,2), \qquad A'(x',2).$ Given point: $B=(2,3), \qquad O=(0,0).$

We are told that the line segments $AB$ and $A'B$ each subtend an angle $\frac{\pi}{4}$ at the origin.
That means the angle between the lines $OA$ and $OB$ is $\frac{\pi}{4}$, and similarly for $OA'$ and $OB$.

So we need all points $A(x,2)$ such that $\angle AOB=\frac{\pi}{4}.$

2. Use slope formula for angle between two lines

Slope of $OB$: $m_1=\frac{3}{2}.$

Slope of $OA$: $m_2=\frac{2}{x}.$

Angle between two lines with slopes $m_1,m_2$ is given by $\tan\theta=\left|\frac{m_2-m_1}{1+m_1m_2}\right|.$

Here $\theta=\frac{\pi}{4}$, so $\tan\theta=1$. Therefore, $\left|\frac{\frac{2}{x}-\frac{3}{2}}{1+\frac{3}{2}\cdot\frac{2}{x}}\right|=1.$

Simplify: $\left|\frac{\frac{2}{x}-\frac{3}{2}}{1+\frac{3}{x}}\right|=1.$

Multiply numerator and denominator by $2x$: $\left|\frac{4-3x}{2x+6}\right|=1.$

So, $|4-3x|=|2x+6|.$

3. Solve the equation

This gives two cases:

Case 1

$4-3x=2x+6$ $-5x=2$ $x=-\frac{2}{5}.$

Case 2

$4-3x=-(2x+6)$ $4-3x=-2x-6$ $x=10.$

Thus the two points are $A\left(-\frac{2}{5},2\right), \qquad A'(10,2).$

4. Find the distance between $A$ and $A'$

Since both points lie on the horizontal line $y=2$, their distance is just the difference of their $x$-coordinates: $AA'=\left|10-\left(-\frac{2}{5}\right)\right|=10+\frac{2}{5}=\frac{52}{5}.$

5. Compare with options

$\frac{52}{5}$ corresponds to Option C.

6. Verification with stored answer

Stored correct answer: C
Derived answer: C

They agree.