Should All the red point pass by a white's circle ?

Simply choose $10$ points such that no four points lie on a circle and colour them white. Now the number of red points is at most $\binom{10}{3} = 120>100$. If number of white points is less than $10$, then red points is less than $\binom{9}{3} = 84$. But $84+9<100$, the answer is $\boxed{10}$.

@TDP I think this question is too easy because I know you are not a fool!

Can we put all the red points in one circle? If it will be true we need only 3 white points to put all the 97 red points in one circle.

A.Amoozadeh wrote: Can we put all the red points in one circle? If it will be true we need only 3 white points to put all the 97 red points in one circle. I guess you read the question incorrectly. It says that the red points are centers of the circle passing through three or more white points.

As we have a unique circle passing through 3 points, we hence have a unique red point for each 3 unique combination of 3 white points. Hence, the no. of unique circles should be greater than the no of red points $\implies\binom{n}{3}\geq100-n$. We see that the smallest no. satisfying is $\boxed{10}$