OK, so I\'ve written most of a program that will allow me to determine if two circles overlap.
I have no problems whatsoever with my program aside from one issue: t
Based on the @trashgod's comment, this is the simpliest way to calculate >distance:
double distance = Math.hypot(x1-x2, y1-y2);
From documentation of Math.hypot:Returns:
sqrt(x²+ y²)
without intermediate overflow or underflow.Bob
Below Bob's approved comment he said he couldn't explain what the
Math.hypot(x1-x2, y1-y2);
did. To explain a triangle has three sides. With two points you can find the length of those points based on the x,y
of each. Xa=0, Ya=0
If thinking in Cartesian coordinates that is (0,0)
and then Xb=5, Yb=9
Again, cartesian coordinates is (5,9)
. So if you were to plot those on a grid, the distance from from x to another x assuming they are on the same y axis is +5
. and the distance along the Y axis from one to another assuming they are on the same x-axis is +9
. (think number line) Thus one side of the triangle's length is 5, another side is 9. A hypotenuse is
(x^2) + (y^2) = Hypotenuse^2
which is the length of the remaining side of a triangle. Thus being quite the same as a standard distance formula where
Sqrt of (x1-x2)^2 + (y1-y2)^2 = distance
because if you do away with the sqrt on the lefthand side of the operation and instead make distance^2 then you still have to get the sqrt from the distance. So the distance formula is the Pythagorean theorem but in a way that teachers can call it something different to confuse people.