Decision boundary in Logistic regression? According to me in logistic regression, we just try to get a line (polynomial)
and then based on which side the point is from that line the sigmoid gives ans>=0.5 or <0.5,
which we then interpret as 1 (if ans>=0.5) and 0 if (ans < 0.5).
Here is a simple plot what I mean.

This is just a random line not a line predicted by any logistic regression model.
Here we are only having one feature i.e., x. y is the output(0 or 1).
The decision boundary here is x = -3 and x = 3.
So when we have 1 feature we try to get a good line to predict the values.
What if we have 2 features we try to get a plane instead?
This is a 3-d plot of a normal sigmoid.

So if we have 2 features say x and y.
For these we are trying to get a plane in z axis which separate the points? which is in turn the decision boundary.
I am really confused.? Is this what we are doing?
If yes, can someone plot a 3d plot (of sigmoid/decision boundary) for a circle.
For eg:

This is also a random plot. Suppose that our points are like this and our model has given us a circle to differentiate between 0 and 1. So can anyone plot a decision boundary of something like this (which is is 3-D).
 A: The sigmoid function takes the linear predictor of the logistic regression and bends it into a range, $[0,1]$, that can interpreted as a probability. The linear predictor can take any real number, which is fine if we think about the log-odds. However, for valid probability values, the must be a transformation to restrict outputs to $[0,1]$.
Any business about a decision boundary is separate from the sigmoid function. In fact, it is totally reasonable to do a decision boundary in terms of log-odds and skip the sigmoid function altogether. Just evaluate if the prediction has positive or negative log-odds, and classify accordingly. This is equivalent to the decision rule in the OP, and it is a worthwhile exercise to consider why.
As always, hard classifications need not be made, since a logistic regression explicitly predicts probability values that can be valuable on their own without considering hard classifications (at least not initially). Our Stephen Kolassa’s answer here gets into some of that.
