Is point exist in circle? Let us consider x and y is a point and then make a radius of some value r.If suppose i had a point a and b, I need to check whether the point a and b is in the radius r from x and y or not. Can give me the formula to this solution,So that i can implement in c programming. 
 A: The point $(a,b)$ is inside the circle of radius $r$ around $(x,y)$ iff $(a-x)^2+(b-y)^2<r^2$.
A: You can easily get rid of the square root as well.
Here is the distance formula:$$
d = \sqrt{(x - a)^2 + (y - b)^2}
$$
The point is in the circle, if $$d \leq r$$
So your formula is $$
\sqrt{(x - a)^2 + (y - b)^2} \leq r
$$
Just square it and you get $$
(x - a)^2 + (y - b)^2 \leq r^2$$
Since you can' t have a negative radius.
This is useful when you have lots of points, because you can get rid all of the square roots in trade of a single calculation of $r^2$. This is of course good only when you iterate over the circles, and check each point for each circle. Pseudocode:
for x, y, radius in circles:
    rs = radius * radius
    for a, b in points:
        if (x - a)^2 + (y - b)^2 <= rs:
            // The point is in the circle

If you have only a single point, the simple distance formula should be perfect too.
If you don't consider points on the radius of the circle to be in the circle, your formula is $d < r$. (By the way $d = r$ is the formula for point is exactly on the radius)
