A small robot travels over a flat floor…

Step 1. Shift. The lines cross at $(2, 1)$. With $x = X + 2$ and $y = Y + 1$ we get $\dfrac{dY}{dX} = \dfrac{X + Y}{X - Y}$.

Step 2. Solve. With $Y = VX$: $X\dfrac{dV}{dX} = \dfrac{1 + V^2}{1 - V}$. Separating gives $\tan^{-1}V - \tfrac{1}{2}\ln(1 + V^2) = \ln|X| + K$. This is the stated form with $C = K$.

Step 3. The constant. At $(3, 2)$ we have $X = 1$ and $Y = 1$, so $\tan^{-1}1 = \dfrac{\pi}{4}$ and the log term is $\tfrac{1}{2}\ln 2$. Then $C = \dfrac{\pi}{4} - \tfrac{1}{2}\ln 2 = 0.4388$.

Step 4. Polar form. With $X = r\cos\theta$ and $Y = r\sin\theta$ the relation says $\theta = \ln r + C$, so $r = e^{\theta - C}$. The same law holds in every direction, because the equation is homogeneous.

Step 5. Evaluate. Along the positive $x$ direction $\theta = 0$, so $r = e^{-0.4388} = 0.6448\ \text{m}$. Along the positive $y$ direction $\theta = \dfrac{\pi}{2}$, so $r = e^{1.5708 - 0.4388} = 3.1018\ \text{m}$.