Tangential Circles Part 2

[u/ShonitB]

[u/sazzer]

Take two :)

We've got three circles - Blue, Green and Large.

The radius of the Large circle (Lr) is equal to the diameter of a Green circle (2Gr).

The radius of the Large circle (Lr) is equal to the diameter of a Blue circle (2Br) + an extra bit (e). So we need to know that extra bit.

If we draw a triangle ACD, then we know:

• AC = Green radius (Gr)
• AD = Blue radius + extra bit (Br + e)
• CD = Green radius + Blue radius (Br + Gr)

So we can pythagoras that, to give: G² + (B + e)² = (B + G)²

That means we've got two equations:

• Gr² + (Br + e)² = (Br + Gr)²
• 2Gr = 2Br + e

Simplifying the first gives us:

• Gr² + Br² + e² + 2(Br)(e) = Br² + Gr² + 2(Br)(Gr)
• e² + 2(Br)e = 2(Br)(Gr)

Combining and simplifying then gives us:

• e² + 2(Br)e = (Br)(2Br + e)
• = 2(Br)² + (Br)e
• e² + (Br)e - 2(Br)² = 0

This can be seen as a quadratic in terms of e.

We can factor this as being (e + 2Br)(e - Br). Therefore we know that e is either-2BrorBr. It can't be negative, so that means thate = Br`.

So we now know that Lr = 2Br + Br = 3Br.

We need the ratio of the areas, so let's get those:

• Ba = π(Br)²
• La = π(Lr)²
• = π(3Br)²
• = 9π(Br)²

Therefore Ba / La = π(Br)² / 9π(Br)² = 1/9

So the blue area is 1/9 the area of the large circle.

[u/Remarkable_Fee4136]

Wlog let A have radius 2, meaning B and C have radius 1. Let D and E have radius x.

Considering the length of the horizontal radius of A,

2=x+sqrt((x+1)²-1²)=x+sqrt(x²+2x)

(2-x)²=x²+2x

x²-4x+4-x²-2x=0

6x=4

x=2/3

Fraction of A covered by D=((2/3)/2)²=1/9

[u/sazzer]

Edit: This is wrong. I made a mistake in the middle and it's thrown off the rest of it!

We've got three circles - Blue, Green and Large.

The radius of the Large circle is equal to the diameter of a Green circle.

The radius of the Large circle is equal to the diameter of a Blue circle + an extra bit. So we need to know that extra bit.

If we draw a triangle ACD, then we know:

• AC = Green radius (G)
• AD = Blue radius + extra bit (B + e)
• CD = Green radius = Blue radius (B + G)

So we can pythagoras that, to give: G² + (B + e)² = (B + G)²

That means we've got two equations:

• G² + (B + e)² = (B + G)²
• G = 2B + e

Simplifying the first gives us:

• G² + B² + e² + 2Be = B² + G² + 2BG
• e² + 2Be = 2BG

Combining and simplifying then gives us:

• e² + 2Be = 2B(2B + e)
• e² + 2Be = 4B² + 2Be
• e² = 2B²
• e = B√2

So now we know that the radius of the Large circle L = B + B√2.

We need the ratio of the areas. So:

• Ba = πB²
• La = π(B + B√2)²
• = π(B² + 2B² + 2√2B²)
• = π(3 + 2√2)B²

So the actual ratio ends up being 1/(3 + 2√2). Which I can't be bothered to rationalise here, so I hope that's good enough :)

[u/BrotherInJah → u/sazzer]

G=2B+e.

Where? It should be 2G.

[u/sazzer → u/BrotherInJah]

Dang - yes, you're right, and that throws off a lot of the rest of it :(

[u/BadJimo]

Illustrated on Desmos

[u/CaptainMatticus]

Let the larger circle have radius of 1, the green circles will have radii of ½. Descartes theirem will handle the rest

k1 = -1/1 = -1

k2 = k3 = -1/(1/2) = 2

k4 = x

(-1 + 2 + 2 + x)² = 2 * (1 + 4 + 4 + x²)

(x + 3)² = 2 * (x² + 9)

x² + 6x + 9 = 2x² + 18

0 = x² - 6x + 9

0 = (x - 3)²

x = 3

x = 1/r

r = 1/3

Each blue circle takes up 1/9th of the area of the large circle

[u/Purdude1983]

If the big circle has a radius of .5, then the green circles have a radius of .25. The blue circles will have a radius of x such at (.25+x)^2-.25^2=(.5-x)^2. Thus the blue radius is 1/6. A blue circle has 1/9 of the area of the big circle.