How to find if a point is outside a circle circumference area? I'd like to know if it's possible to calculate if a point is inside or outside the circle circumference area based on it's $x$ and $y$ values ? Example, $(x, y)= (0.85, -0.9)$ and the radius is $1$

 A: You need only calculate the distance between $(x_1, y_1)$ (the point in question) and $(x_0, y_0)$ (your centre) using the standard distance formula $$d = \sqrt{(x_1 - x_0)^2 + (y_1 - y_0)^2}$$


*

*If $d = 1$, then the point lies on the circumference. 

*If $d > 1$, then the point is outside the circle. 

*If $d< 1$, then the point is inside the circle. 

A: Use Pythagorean's Theorem!
If $x^2 + y^2 = r^2$, where $x$ and $y$ are the coordinates in question, then every point inside of the circle is given by the solutions of $x^2 + y^2 \leq r^2$ (assuming the center of your circle is at the origin).
A: Find the distance between the point $(x,y)$ and the center of the circle.
Consider your point is $(x,y)$ and the center of the circle is $(x_1,y_1)$ and $r$ is the radius of the circle.
If $d = (x-x_1)^2 + (y-y_1)^2 > r^2$ then the point is outside the circle.
If $d<r^2$ then the point is inside and if $d = r^2$ then the point is on circumstance of the circle.
A: Notice, 
in general, equation of the circle having center $(x_1, y_1)$ & radius $r$ unit is given as $$(x-x_1)^2+(y-y_1)^2=r^2$$ 
Let there be any arbitrary point say $(x_o, y_o)$ then if 


*

*$\sqrt{(x_o-x_1)^2+(y_o-y_1)^2}>r$ the point lies outside the circle 

*$\sqrt{(x_o-x_1)^2+(y_o-y_1)^2}=r$ the point lies on the circumference of circle 

*$\sqrt{(x_o-x_1)^2+(y_o-y_1)^2}<r$ the point lies inside the circle 


For example, a given point is $(0.85, -0.9)$ & circle is centered at the origin $(0, 0)$ having a radius $1$ unit. The distance of the point from the center of the circle is $$\sqrt{(0.85-0)^2+(-0.9-0)^2}\approx 1.23794184>1\ \text{(radius of circle)}$$ Thus, the point $(0.85, -0.9)$ lies out side the circle: $x^2+y^2=1$
