coordinate systemsgeometry
So let's say that I have a standard $(x, y)$ coordinates system, the center of a circle $(0,1)$ and the radius $(1)$. How can I determine that whether the point $(0,0)$ touches the circumference or not?
I don't need to determine if the point is inside or outside the circumference but my problem is related with its touching the circumference of the following circle.
Let C be the circle located inside the first quadrant. (The logic will be the same if the origin is located at the top left corner.) We let its equation be $F(x, y) = (x – h)^2 + (y – k)^2 - r^2 = 0$.
Initial task == Testing whether the mouse pointer (at T(p, q)) is inside or outside of the circle.
This is easily done by testing whether F(p, q) is greater than or equal to or smaller than 0. The conclusion will be outside/ on/ inside the circle accordingly.
We assume that T(p, q) is inside the circle.
The corresponding point (the one on the circumference) is then (p, m) where m will have 2 values (? and ??).
From the fact that $m – k = \pm \sqrt(r^2 – (p – h)^2)$, and with the help of the diagram, we conclude that $?? = k + \sqrt(r^2 – (p – h)^2)$.
The answer should be the circle touching y=|x| and y=4-$x^2$ from above with r= 4(1+$\sqrt2$). Its radius is larger than 4($\sqrt2$-1), which is the radius of the other circle touching the parabola from underneath.
As shown in the graph, let r= radius of the circle centered at o'(0,x+r) , the circle touches y=x at A(x,x).
With $\Delta$ OO'A, $(4+r)^2$=$r^2$+$(\sqrt2x)^2$
With $\Delta$ BO'A, $(4+r-x)^2+x^2$=$r^2$
Solving these two equations,
x($x^2-8x+8)=0, $r=$\frac{x^2-8}{4}$
Only solution x=4+2$\sqrt2$, r=4(1+$\sqrt2$) provides positive r.
Best Answer
The general equation for the circle is $(x-c_x)^2 + (y-c_y)^2 = r^2$, where $(c_x,c_y)$ is the center and $r$ is the radius. A point $(x,y)$ lies on the circle (the disk's boundary) if it satisfies the above equation. If instead: $(x-c_x)^2 + (y-c_y)^2 < r^2$ then it is inside the disk, and similarly it is outside for $>$.
Edit: This can be extended to arbitrary high dimensions, let $\vec{c}\in \mathbb{R}^n$ be the center of the $n$ dimensional hypersphere, and $r$ be its radius. Then a point $\vec{v}\in \mathbb{R}^n$ lies on its surface if $|\vec{v}-\vec{c}|^2 = r^2$, where $|\vec{a}| = \sqrt{\sum_{i=1}^{n}{a_i^2}}$ is the Euclidean norm. It is inside the hyperball for $<$ and outside for $>$.