# Take a sphere and draw on it a great circle (a great circle is a circle whose centre is the centre o

Phoebe Xiong 2022-03-03 Answered
Take a sphere and draw on it a great circle (a great circle is a circle whose centre is the centre of the sphere). There are two regions created. Here, I am referring to regions on the surface of the sphere. Now draw another great circle: there are four regions. Now draw a third, not passing through the points of intersection of the first two. How many regions?
Here's the general question: How many regions are created by n great circles, no three concurrent, drawn on the surface of the sphere?
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## Expert Answer

Tail3vn
Answered 2022-03-04 Author has 3 answers

Step 1
Calculate the number of the regions for 3 non - collinear circles.
For 1 great circle $=2$ regions
For 2 great circles $=4$ regions
For 3 great circles $=8$ regions
As the number of the circles increase by 1, the number of regions increase by 2n times.
Therefore, for ${\left(n+1\right)}^{th}$ circle the number of the regions increse by 2n times.
The sequence is a quadratic sequence.
Step 2
Determine the quadratic sequence.
Sequence for number of regions $=2,4,8,14\cdots$
${T}_{n}=a{n}^{2}+bn+c$
${T}_{1}=2$
$a{\left(1\right)}^{2}+b\left(1\right)+c=2$
$a+b+c=2$ (1)
${T}_{2}=4$
$a{\left(2\right)}^{2}+b\left(2\right)+c=4$
$4a+2b+c=4$ (2)
${T}_{3}=8$
$a{\left(3\right)}^{2}+b\left(3\right)+c=8$
$9a+3b+c=8$ (3)
Step 3
After solving the 3 equations, the values for and $c=2$.
${T}_{n}=a{n}^{2}+bn+c$
and $c=2$
${T}_{n}=\left(1\right){n}^{2}+\left(-1\right)n+2$
${T}_{n}={n}^{2}-n+2$
Hence, the number of regions for the n circles is
${T}_{n}={n}^{2}-n+2$
where, n is the number of circleswhich are non collinear.

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