Since you seem to know the answer, I will give it here.

Suppose that $v_l = v_m / k $ and the radius of the lake is $r$. Then the lady can reach a distance $\frac{r}{k}$ from the centre and keep the monster directly behind her, a distance $r\left(1 + \frac{1}{k}\right)$ away. One way would be to swim in a spiral gradually edging outwards as the monster runs trying to close the distance; another would be to swim in a semi-circle of radius $\frac{r}{2k}$ away from the monster once it starts to run. And the lady can sustain this distance by going round in a circle as the monster tries in vain to close the distance.

The next stage is for the lady to try to swim direct to shore at some point away from the direction the monster is running. If the monster starts at the point $(-r,0)$ running anti-clockwise and the lady starts at the point $\left(\frac{r}{k},0\right)$ her best strategy is to head off in a straight line initially at right angles to the line between her and the monster: a less steep angle and the monster has proportionately less far to run than the lady has to swim, but a steeper angle and it is worth the monster changing direction. (If the monster changes direction in this right-angle case, the lady changes too but now starts closer to shore.) As they are both trying to get to the point $\left(\frac{r}{k},r \sqrt{1-\frac{1}{k^2}}\right)$ then they will arrive at the same time if $ \pi + \cos^{-1}(1/k) = k \sqrt{1 -1/k^2}$ which by numerical methods gives $k \approx 4.0633$.

So if the monster is less than 4.0633 times as fast as the lady, the lady can escape; if not then she stays in the lake and the monster stays on the edge and they live unhappily ever after.

Since you seem to know the answer, I will give it here.

Suppose that $v_l = v_m / k $ and the radius of the lake is $r$. Then the lady can reach a distance $\frac{r}{k}$ from the centre and keep the monster directly behind her, a distance $r\left(1 + \frac{1}{k}\right)$ away. One way would be to swim in a spiral gradually edging outwards as the monster runs trying to close the distance; another would be to swim in a semi-circle of radius $\frac{r}{2k}$ away from the monster once it starts to run. And the lady can sustain this distance by going round in a circle as the monster tries in vain to close the distance.

The next stage is for the lady to try to swim direct to shore at some point away from the direction the monster is running. If the monster starts at the point $(-r,0)$ running anti-clockwise and the lady starts at the point $\left(\frac{r}{k},0\right)$ her best strategy is to head off in a straight line initially at right angles to the line between her and the monster: a less steep angle and the monster has proportionately less far to run than the lady has to swim, but a steeper angle and it is worth the monster changing direction. (If the monster changes direction in this right-angle case, the lady changes too but now starts closer to shore.) As they are both trying to get to the point $\left(\frac{r}{k},r \sqrt{1-\frac{1}{k^2}}\right)$ then they will arrive at the same time if $ \pi + \cos^{-1}(1/k) = k \sqrt{1 -1/k^2}$ which by numerical methods gives $k \approx 4.6033$.

So if the monster is less than 4.6033 times as fast as the lady, the lady can escape; if not then she stays in the lake and the monster stays on the edge and they live unhappily ever after.