What is the distance of the tangent between two circles

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The drawing is a bit misleading. The radii drawn to the tangent point in both circles are perpendicular to the same tangent line, so they must be parallel. Do you know how to draw the common tangent to two circles? You shrink both circles till the smaller one becomes a point, then the problem reduces to find the length of the tangent line drawn from the centre of the smaller circle.

ehild
 
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so if both circles have the same radius then the distance between the radaii is the length of the tangent right?
 
and also let's say i have multiple points refer the pic
and the points are
x y
1 4
3 2
7 9
5 4
9 5
6 7
9 1
11 8

the radius is 1 unit

i did a convex hull algorithm and found the perimeter and added the arc length. since it is a 5 sided polygon the interior angles are 540 degrees

the total perimeter is 25.20983226924521 units and with the arc (9.4247779607694 units) its equal to 34.63461023001461 units but the actual solution was 34.408 units (rounded).
so what am i doing wrong?

u can find this at http://wcipeg.com/problem/boi09p1
 
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BloodyFrozen said:
Did you make an attempt at the problem?
yes i did look above ^^
 
I don't know anything about your convex hull algorithm. But the total of the arcs is just the circumference of one mine isn't it? If the radius is 1, isn't that just ##2\pi##? How are you getting 9.424...?
 
LCKurtz said:
I don't know anything about your convex hull algorithm. But the total of the arcs is just the circumference of one mine isn't it? If the radius is 1, isn't that just ##2\pi##? How are you getting 9.424...?

in the picture there are 5 mines that stick out. so the internal angles is 540 degrees so the length of the arc. rather than 360 degrees this is 540 degrees that's how i got ~ 9.424
 
LCKurtz said:
I don't know anything about your convex hull algorithm. But the total of the arcs is just the circumference of one mine isn't it? If the radius is 1, isn't that just ##2\pi##? How are you getting 9.424...?

poioip said:
in the picture there are 5 mines that stick out. so the internal angles is 540 degrees so the length of the arc. rather than 360 degrees this is 540 degrees that's how i got ~ 9.424

I think you should give that some more thought.
 
LCKurtz said:
I think you should give that some more thought.

i do not understand what's wrong
so if i draw tangents on the outside it will join and form a pentagon and the internal angles are 540. so what am i doing wrong? please help me!
 
poioip said:
i do not understand what's wrong
so if i draw tangents on the outside it will join and form a pentagon and the internal angles are 540. so what am i doing wrong? please help me!

The internal angles don't have anything to do with the arcs. Look at just the arcs and imagine putting them together ignoring the straight sections.
 
i see it forms a single circle!

if the perimeter is (25.20983226924521) + 2*3.1415926535897932384626433832795 is
31.493017576424796476925286766559 not even close to 34.408
 
poioip said:
i see it forms a single circle!

if the perimeter is (25.20983226924521) + 2*3.1415926535897932384626433832795 is
31.493017576424796476925286766559 not even close to 34.408

What I would do if I were you is work the sample problem out manually without using your convex hull algorithm. Plot the points calculate the convex hull distances directly. That way you will have a check on whether your answer or their answer is correct. And if it turns out theirs is, it's time to turn your attention to your convex hull algorithm. Good luck with it.
 
LCKurtz said:
What I would do if I were you is work the sample problem out manually without using your convex hull algorithm. Plot the points calculate the convex hull distances directly. That way you will have a check on whether your answer or their answer is correct. And if it turns out theirs is, it's time to turn your attention to your convex hull algorithm. Good luck with it.

i did check it manually and i was wrong.the algorithm is just to identify the points that stick out!
 
I get the perimeter ≈ 28.12465 + 2πR ≈ 34.40784
 
Whether COSINE^4(x) is a periodic function or not ?