+118673 Here is a fun question for some people.
If it is very easy for you then please do not answer.
It is intended to be a interesting challenge question.
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Assume the Earth is a perfect sphere.
A peice of string is wrapped around the equator to make a snug fit.
Now another string exactly one metre longer than the first one is wrapped around the earth so that the 2 peices of string form 2 concentric circles.
How far off the surface of Eath is the second peices of string?
+2863 \(2{\pi}r=C\)
That is the formula for circumference.
Now to find how far off, we find the difference between the two radii when the concentric circles are formed.
We make a system of equations
\(2{\pi}x=C\)
\(2{\pi}y=(C+1)\)
We have to evaluate \(|y-x|\)
Substituting: \(2{\pi}y-1=2{\pi}x\)
Simplifying: \(1=2{\pi}y-2{\pi}x\)
Solving: \(\frac{1}{2{\pi}}=y-x\)
Solving for \(|y-x|\),
\(\boxed{\frac{1}{2{\pi}}}\)
That was my attempt 
+2863 \(2{\pi}r=C\)
That is the formula for circumference.
Now to find how far off, we find the difference between the two radii when the concentric circles are formed.
We make a system of equations
\(2{\pi}x=C\)
\(2{\pi}y=(C+1)\)
We have to evaluate \(|y-x|\)
Substituting: \(2{\pi}y-1=2{\pi}x\)
Simplifying: \(1=2{\pi}y-2{\pi}x\)
Solving: \(\frac{1}{2{\pi}}=y-x\)
Solving for \(|y-x|\),
\(\boxed{\frac{1}{2{\pi}}}\)
That was my attempt
+118673 Great work CalculatorUser! 
You maths is excellent,
Just for a slightly different presentation I could say:
\(C=2\pi r\\ \frac{C}{2\pi}=r\\ \frac{C+1}{2\pi}=\frac{C}{2\pi}+\frac{1}{2\pi}=r+\frac{1}{2\pi} \)
So if the circumference is increased by 1 unit, the radius is increased by \(\frac{1}{2\pi}\;units\)
So if the circumference is increased by 1 metre, the radius is increased by \(\frac{1}{2\pi}\;metre\)
And this equals an approximate radius increas of 0.159 metres or approx 16cm
This distance between the 2 concentric circles will always be the same.
It does not matter how big or small the original circle is.
