At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

See more answers at brainly.com

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the **expert** answer you'll need to create a **free** account at **Brainly**

given 3 circles, the circumference of the 2nd circle cuts the centre of circle1 at O1, while the centre of the 3rd circle is formed between the upper intersection of 1st and 2nd circle, with its circumference cuts through centre of 1st circle.
radius of 2nd circle is unknown.

O1-P to O1-O2 are these distances

yep

o1 to p i the base of an iso tri with sides of r|dw:1344615311196:dw|

we can define an angle using o1,o2 and oi,1 so that o1,p is findable to me

err ... o1,o3 that is

|dw:1344615460417:dw|

or really P = (2a,0) in this case :)

|dw:1344615575595:dw|
im sure we could come up with something

we can determine a general ratio; and when given specifics we can fill them in

so i will need to come out with 3 ratios for O1P < O1O2, O1P = O1O2, and O1P > O1O2?

because radius for circle 2 is not fixed

radius for circle 2 is "fixed"

owh. k i'll try from here. thx =)

i might be reading the post wrong tho

o2 is solid, so do we assume we know the position of o2 relative to the other solid objects?

erm o2 is the centre of 2nd circle. lol i drawed the dot too big i guess.

its an imaginary centre because radius 2 is unknown.

but centre of circle 2 will be placed same level with circle 1

|dw:1344616423539:dw|
well, you do have 3 points of contact for the circle2 i believe

|dw:1344616612225:dw|

|dw:1344616835067:dw|
y^2 = x*c ; x+c = radius from 01 to 02

or rather r' = x+c

the ratio then is.... 2x : r'

2x : x+c , and since y^2 = xc; c=y^2/x
2x : x + y^2/x
2x : (x^2 + y^2)/x

o i'll try to digest 'em. really thx for spending time with this question. @@

youre welcome, it was fun :)

y^2 = xc is some theorem? o.o