Position of a point relative to a circle intuitively.

Question

I'm studying co-ordinate geometry and would like to intuitively (well, graphically actually) understand the statement

If we put the the values of $x$ and $y$ co-ordinates of a point in the expression $x^2 + y^2 - r^2$ we get a +ve value if a point lies outside, -ve value if a line lies inside of the circle $x^2 + y^2 - r^2 = 0$ .

I cannot justify to myself as why the exterior part should be +ve and interior -ve.
I can prove the statement by algebra (by comparing distance of point with radius) but I want something more intuitive, more obvious.
Assume I don't know the distance formula. I know I might be sounding weird, sorry for that.

PS: Homework I've done:
To prove:

When a point $(x,y)$ is put in expression $ax+by+c$ the positive and negative value tells about the side-ness of the point relative to the line $ax+by+c=0$

Proof: Consider the function $f(x,y) =x+y$ (It can be $ax+by+c$ , makes no difference, the nature of the graph would be same)

Source: https://academo.org/demos/3d-surface-plotter/?expression=x%2By&xRange=-50%2C+50&yRange=-50%2C+50&resolution=25

This clearly shows that the point travelling from one side of the line $x+y=0$ to other must change it's sign when put in the expression $x+y$

PPS: Please provide a better explanation, if possible, to the position of point w.r.t. line as well.

PPPS : Can there be a general way to think like this for all conic sections ?

Answer 1

Without loss of generality we can assume as centre the origin $O(0,0)$

Any point of the plane $P(x,y)$ has a distance from $O$ such that $OP^2=x^2+y^2$

If the point is on the circumference having radius $r$ then $x^2+y^2=r^2$ so $x^2+y^2-r^2=0$

If it is inside the disk then $x^2+y^2<r^2$ you have $x^2+y^2-r^2<0$

and if it is outside then $x^2+y^2>r^2$ that is $x^2+y^2-r^2>0$

Hope this helps