# PHP Exercises: Test if circumference of two circles intersect or overlap

## PHP: Exercise-58 with Solution

There are two circles C1 with radius r1, central coordinate (x1, y1) and C2 with radius r2 and central coordinate (x2, y2)

Write a PHP program to test the followings -

"C2 is in C1" if C2 is in C1

"C1 is in C2" if C1 is in C2

"Circumference of C1 and C2 intersect" if circumference of C1 and C2 intersect, and and
"C1 and C2 do not overlap" if C1 and C2 do not overlap.

**Input:**Input numbers (real numbers) are separated by a space.

Input 0 to exit.

**Sample Solution: **-

**PHP Code:**

```
<?php
$n = intval(fgets(STDIN));
for($i = 0; $i < $n; $i++){
fscanf(STDIN, "%lf %lf %lf %lf %lf %lf", $xa, $ya, $ra, $xb, $yb, $rb);
$r = sqrt(($xb - $xa)*($xb - $xa) + ($yb - $ya)*($yb - $ya));
if($r + $ra < $rb){
echo "C1 is in C2\n";
continue;
}
if($r + $rb < $ra){
echo "C2 is in C1.\n";
continue;
}
if($r <= $ra + $rb){
echo "Circumference of C1 and C2 intersect.";
continue;
}
echo "C1 and C2 do not overlap.\n";
}
?>
```

**Sample Input:**

2

0.0 0.0 6.0 0.0 0.0 5.0

0.0 0.0 3.0 5.1 0.0 3.0

Sample Output:

C2 is in C1. Circumference of C1 and C2 intersect.

**Flowchart: **

**PHP Code Editor:**

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**Previous:** Write a PHP program to find the maximum sum of a contiguous subsequence from a given sequence of numbers a1, a2, a3, ... an. A subsequence of one element is also a continuous subsequence.

**Next:** Write a PHP program to that reads a date (from 2004/1/1 to 2004/12/31) and prints the day of the date. Jan. 1, 2004, is Friday. Note that 2004 is a leap year.

**What is the difficulty level of this exercise?**

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## PHP: Tips of the Day

**PHP: How to generate a random, unique, alphanumeric string for use in a secret link?**

**Security Notice:** This solution should not be used in situations where the quality of your randomness can affect the security of an application. In particular, rand() and uniqid() are not cryptographically secure random number generators. See Scott's answer for a secure alternative.

If you do not need it to be absolutely unique over time:

md5(uniqid(rand(), true))

Otherwise (given you have already determined a unique login for your user):

md5(uniqid($your_user_login, true))

Ref : https://bit.ly/31fd9wa

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