Circles of given radius through two points: Difference between revisions
Circles of given radius through two points (view source)
Revision as of 00:11, 22 May 2020
, 4 years ago→{{header|Haskell}}
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Center: {0.1234 0.9876}
</pre>
=={{header|Groovy}}==
{{trans|Java}}
<lang groovy>class Circles {
private static class Point {
private final double x, y
Point(Double x, Double y) {
this.x = x
this.y = y
}
double distanceFrom(Point other) {
double dx = x - other.x
double dy = y - other.y
return Math.sqrt(dx * dx + dy * dy)
}
@Override
boolean equals(Object other) {
//if (this == other) return true
if (other == null || getClass() != other.getClass()) return false
Point point = (Point) other
return x == point.x && y == point.y
}
@Override
String toString() {
return String.format("(%.4f, %.4f)", x, y)
}
}
private static Point[] findCircles(Point p1, Point p2, double r) {
if (r < 0.0) throw new IllegalArgumentException("the radius can't be negative")
if (r == 0.0.toDouble() && p1 != p2) throw new IllegalArgumentException("no circles can ever be drawn")
if (r == 0.0.toDouble()) return [p1, p1]
if (Objects.equals(p1, p2)) throw new IllegalArgumentException("an infinite number of circles can be drawn")
double distance = p1.distanceFrom(p2)
double diameter = 2.0 * r
if (distance > diameter) throw new IllegalArgumentException("the points are too far apart to draw a circle")
Point center = new Point((p1.x + p2.x) / 2.0, (p1.y + p2.y) / 2.0)
if (distance == diameter) return [center, center]
double mirrorDistance = Math.sqrt(r * r - distance * distance / 4.0)
double dx = (p2.x - p1.x) * mirrorDistance / distance
double dy = (p2.y - p1.y) * mirrorDistance / distance
return [
new Point(center.x - dy, center.y + dx),
new Point(center.x + dy, center.y - dx)
]
}
static void main(String[] args) {
Point[] p = [
new Point(0.1234, 0.9876),
new Point(0.8765, 0.2345),
new Point(0.0000, 2.0000),
new Point(0.0000, 0.0000)
]
Point[][] points = [
[p[0], p[1]],
[p[2], p[3]],
[p[0], p[0]],
[p[0], p[1]],
[p[0], p[0]],
]
double[] radii = [2.0, 1.0, 2.0, 0.5, 0.0]
for (int i = 0; i < radii.length; ++i) {
Point p1 = points[i][0]
Point p2 = points[i][1]
double r = radii[i]
printf("For points %s and %s with radius %f\n", p1, p2, r)
try {
Point[] circles = findCircles(p1, p2, r)
Point c1 = circles[0]
Point c2 = circles[1]
if (Objects.equals(c1, c2)) {
printf("there is just one circle with center at %s\n", c1)
} else {
printf("there are two circles with centers at %s and %s\n", c1, c2)
}
} catch (IllegalArgumentException ex) {
println(ex.getMessage())
}
println()
}
}
}</lang>
{{out}}
<pre>For points (0.1234, 0.9876) and (0.8765, 0.2345) with radius 2.000000
there are two circles with centers at (1.8631, 1.9742) and (-0.8632, -0.7521)
For points (0.0000, 2.0000) and (0.0000, 0.0000) with radius 1.000000
there is just one circle with center at (0.0000, 1.0000)
For points (0.1234, 0.9876) and (0.1234, 0.9876) with radius 2.000000
an infinite number of circles can be drawn
For points (0.1234, 0.9876) and (0.8765, 0.2345) with radius 0.500000
the points are too far apart to draw a circle
For points (0.1234, 0.9876) and (0.1234, 0.9876) with radius 0.000000
there is just one circle with center at (0.1234, 0.9876)</pre>
=={{header|Haskell}}==
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