# Circle Packing

N circles have been packed inside a square of side length R unit and radius of the circle is to be calculated.

# Case 1

**When**

*n*= 1

Obviously in this case, diameter of the circle = side of the square.

So, the radius of the circle R unit

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# Case 2

**When**

*n*= 2Consider this figure

This is a right-angled triangle with sides r, r and r âˆš2. Hence the diagonal of the square = 4r + 2 r âˆš2 = âˆš2 R

# Case 3

**When**

*n*= 4â€‹

Obviously, in this case 2 Ã— diameter of a circle = side of square.

# Case 4

**When**

*n*= 5â€‹

Consider this figure:

If the radius of the circle is r, then the distance beÂtween the centre of the circle and the vertex of a square = r âˆš2

Now, consider this figure, the triangle formed here is a right-angled triangle:

1/âˆš2(1/2 of the diagonal of the square) = r + r + r âˆš2
= r(2 + âˆš2), so,

# Case 5

**When**

*n*= 8â€‹

We will find out the value of r in 3 steps here:

*Step 1*Find 1 - 2 = 5 - 6

*Step 2*Find 2 - 3 = 4 - 5

*Step 3*Find 3 - 4

*Step 1*1 - 2 = 5 - 6 = r + r âˆš2

*Step 2*For 2 - 3

So, 2 - 3 = 4 - 5 = râˆš3 â€“ r

Step 3

Step 3

This is a square with side length 2 r and diagonal = 2r âˆš2
So, 3 - 4 = 2r
Hence 1 â€“ 6 = 2(r + râˆš2) + 2(râˆš3 â€“ r) + 2r = Râˆš2. Now r can be calculated.

# Case 6

**When**

*n*= 9â€‹

Radius of circle = r
So, the side of a square = 3 Ã— diameter of a circle = 6r
So, 6r = R unit, r = R/6 unit

# Circles Inside Circles

Radius of the outer circle R = 1 unit and n similar circles of â€˜râ€™ radius have been inserted inside this outer circle.

N = 1

Obviously R = r

N = 2

Obviously, 4 r = 2 R Hence r = R/2

N = 3

The triangle formed inside will be an equilateral triangle of the side length 2 r. The centre of this circle will be incentre/centroid/circumcentre of this equilateral triangle. So, the distance between the centre of any smaller circle to the centre of the bigger circle = 2/3 median =

*x*= 2/3 [âˆš3/2 Ã— a], where a = 2r.

*x*= R. Now put the values of

*x*from the above condition to find the value of r.

N = 4

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The square formed inside will be of the side length 2r. Now consider this figure,

N = 5

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The figure inside is a right angled triangle, with base and height being equal to r, and hypotenuse = râˆš2 = r +

*x*. Hence

*x*= r(âˆš2 â€“ 1)

*x*. Hence r = 2(R â€“

*x*)/4[Now put the value of

*x*from the above condition.]

N = 6

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Here we can put a similar circle inside all the six circles making it a 7 circle figure. Now, the situation is the same as that of a circle packing with N = 7
In this case, 6r = R, so, r = 1/6 R

N = 7

In this case, 6r = R, so, r = 1/6 R

# Ptolemyâ€™s Theorem of Cyclic Quadrilateral

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For a cyclic quadrilateral, the sum of the products of the two pair of the opposite sides equal the product of the diagonals.
In the figure given above, AB Ã— CD + BC Ã— DA = AC Ã— BD
In any cyclic quadrilateral of side lengths

This means that a parallelogram drawn inside a circle is always a rectangle.

*a*,*b*,*c*and*d*, the length of the diagonals*p*and*q*will be equal to:Example

Under some special conditions, it is given that a cyclic quadrilateral ABCD is a parallelogram. What kind of figure will ABCD be?

Solution

ABCD is a parallelogram. Thus

*a*+*b*= 180Â°. And âˆ A + âˆ C = 180Â°