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Having a circle with the centre $(x_c, y_c)$ with the radius $r$ how to know whether a point $(x_p, y_p)$ is inside the circle?

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Do you know the formula for the distance between two points, given their coordinates? – Brian M. Scott Sep 18 '12 at 22:32
Well, it would definitely help. – Ivan Sep 18 '12 at 22:34
Inside the radius? You mean inside the circle? – Pedro Tamaroff Sep 18 '12 at 22:35
Exactly, @PeterTamaroff. – Ivan Sep 18 '12 at 22:38
up vote 20 down vote accepted

The distance between $\langle x_c,y_c\rangle$ and $\langle x_p,y_p\rangle$ is given by the Pythagorean theorem as $$d=\sqrt{(x_p-x_c)^2+(y_p-y_c)^2}\;.$$ The point $\langle x_p,y_p\rangle$ is inside the circle if $d<r$, on the circle if $d=r$, and outside the circle if $d>r$. You can save yourself a little work by comparing $d^2$ with $r^2$ instead: the point is inside the circle if $d^2<r^2$, on the circle if $d^2=r^2$, and outside the circle if $d^2>r^2$. Thus, you want to compare the number $(x_p-x_c)^2+(y_p-y_c)^2$ with $r^2$.

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The point is inside the circle if the distance from it to the center is at most $r$. Symbolically, this is $$\sqrt{|x_p-x_c|^2+|y_p-y_c|^2}< r.$$

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I would not say that it’s inside the circle when the distance is $r$. – Brian M. Scott Sep 18 '12 at 22:51
It depends on whether you include the boundary. If you don't, replace $\leq$ with $<$. – Alex Becker Sep 18 '12 at 22:54
I’d say that you’re confusing the (closed) disk with the circle. A point at distance $r$ for the centre is inside the disk but not inside the circle; rather, it’s on the circle. (It’s a fairly minor point, but there are enough occasions for terminological confusion that I like to avoid it when it’s easy to do so.) – Brian M. Scott Sep 18 '12 at 22:57
@BrianM.Scott That's a fair point. – Alex Becker Sep 18 '12 at 22:59

Alex and Brian have answered your question already. In case you're trying to implement this algorithm in some programming language, here's my Haskell implementation:

distance :: Floating a => (a, a) -> (a, a) -> a
distance (x1,y1) (x2,y2) = sqrt((x1-x2)**2 + (y1-y2)**2)

isInsideCircle :: (Ord a, Floating a) => a -> (a, a) -> (a, a) -> Bool
isInsideCircle r (xc,yc) (x,y) | (distance (xc,yc) (x,y) < r) = True
                               | (distance (xc,yc) (x,y) >= r) = False

Suppose you have a circle whose radius is $r = 1$ and whose center is the origin $(0,0)$. You would like to know if $(\frac{1}{2},0)$, $(1,0)$, and $(1,1)$ are inside the circle. The following interactive GHCi session answers the question:

*Main> isInsideCircle 1 (0,0) (0.5,0)
*Main> isInsideCircle 1 (0,0) (1,0)
*Main> isInsideCircle 1 (0,0) (1,1)
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If you have the equation of the circle, simply plug in the x and y from your point (x,y). After working out the problem, check to see whether your added values are greater than, less than, or equal to the r^2 value. If it is greater, then the point lies outside of the circle. If it is less than, the point is inside the circle. If it is equal, the point is on the circle.

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