Six spheres of radius $1$ are positioned so that their centers are at the vertices of a regular hexagon of side length $2$. The six spheres are internally tangent to a larger sphere whose center is the center of the hexagon. An eighth sphere is externally tangent to the six smaller spheres and internally tangent to the larger sphere. What is the radius of this eighth sphere

look in attachments

step-by-step explanation:

a = (-2, 0)

b= (-4, 0)

c = (-5,2)

d = (-1,2)

ιт ιѕ 12 αв

нσρє ι ¢συℓ∂ нєℓρ уα

option (a)

step-by-step explanation:

step 1:   take 2 number whose product equal to 25..

and sum of those two number are -10

step: 2 for example here if we take 5,5 are the number,,

5*5=25

and

5+5=10

but we need -10

so if we take   -5,-5

so -5*-5=25

and -5+(-5)=-10

s0

come the equation

replace -10x by these number

like -5x -5x+25

take common

x(x-5)-5(x-5)=0

(x-5)(x-5)=0