CPhill

Let  the tangent line touch circle A at D   and  circle B  at E

Let the distance from  B to C  = x

We can set up similar triangles ADC  and BEC  such that

AD / AC  =  BE / BC

5 / ( 5 + 3 + x)  =  3  / x

5 ( 8 + x)  = 3   / x          cross-multiply

5x  = 3 (8 + x)

5x  = 24 + 3x

2x  = 24

x = 12  = BC

Impossible

To see why

ACB  = 90°

QCP  = 45°

If the cos ACP  = 2/3....then arcos  (2/3)  ≈ 48.2°

But this means that   QCP + ACP  =  45 + 48.2  =  93.2°   which would be >  ACB

Here :  https://web2.0calc.com/questions/pls-help_13322

Rewrite as

x =  2y^2  - 13y  -1

The y coordinate of the vertex  is    - (-13)  / ( 2*2) = 13/4

The x coordinate is

2(13/4)^2 -13(13/4) - 1  =   

-177/8

Vertex = ( -177/8 , 13/4)

Let the point we want  be    (x , (1/2)(x^2 - 8))

Using the square of the distance

D^2  =  x^2  +   [ (1/2)(x^2 -8) ]^2

Take the derivative and set to 0

2x + (x^2 - 8)(2x)  = 0

2x + 2x^3 - 16x  = 0

2x^3 - 14x  =  0

x^3 - 7x   = 0

x( x^2 - 7)  = 0

x = 0   (reject)

x^2 - 7  = 0

x = sqrt (7)

The point  is  ( sqrt 7 , -1/2)

The distance  is  sqrt  [ (sqrt 7)^2  + (-1/2)^2 ] =  sqrt [ 7 + 1/4 ] =  sqrt (29 ) / 2

The  circumference  of the cone =  2 * pi * 21  * (120/360)   =  14 pi

The radius of the cone = 

14pi =  2 * pi * r

7 = r

Height of the cone  = sqrt [ 21^2 - 7^2 ]  = sqrt (392) = 14sqrt 2

Volume of cone =   (1/3) pi (7)^2  * 14sqrt 2   = 686sqrt (2 ) / 3   pi   cm^3

Formula  for the number of lines, n,  that will divide a plane into k regions

(1/2) (n^2 + n + 2)  =  k

(1/2) (n^2 + n + 2)  =   5

n^2 + n + 2  = 10

n^2 + n =  8

n  must be  greater than 2

Here : https://web2.0calc.com/questions/sphere_18

Nothing shown

Sorry....I thought this problem was similar to another one asked on here earlier.....see my corrected answer ....