- #1

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- Homework Statement
- see attached

- Relevant Equations
- pythagoras theorem

Find the question here and the solution i.e number 10 indicated as ##6-4\sqrt{2}##,

I am getting a different solution, my approach is as follows. I made use of pythagoras theorem for the three right angle triangles as follows,

Let radius of the smaller circle be equal to ##c## and distance from mid point of smaller circle to circle with radius ##1=x## then it follows that,

##x^2+(1-c)^2=(1+c)^2##

##x^2=4c##

let us also have the distance from the smaller circle to circle with radius ##2=y##, then it follows that,

##y^2+(2-c)^2=(2+c)^2##

##y^2=8c##

Let ##y^2+x^2=AB##, then it follows that, ##(AB)^2 + (2-1)^2=3^2##

##AB=2\sqrt{2}##

then it follows that, ##\sqrt{8}=2\sqrt{c}+2\sqrt{2c}##

Am i missing something here!

i got it!

##\sqrt{c}##=##\dfrac {\sqrt {8}}{2+2\sqrt{2}}##

⇒##c=0.34314575## Bingo guys! Phew i took time on this men!

I am getting a different solution, my approach is as follows. I made use of pythagoras theorem for the three right angle triangles as follows,

Let radius of the smaller circle be equal to ##c## and distance from mid point of smaller circle to circle with radius ##1=x## then it follows that,

##x^2+(1-c)^2=(1+c)^2##

##x^2=4c##

let us also have the distance from the smaller circle to circle with radius ##2=y##, then it follows that,

##y^2+(2-c)^2=(2+c)^2##

##y^2=8c##

Let ##y^2+x^2=AB##, then it follows that, ##(AB)^2 + (2-1)^2=3^2##

##AB=2\sqrt{2}##

then it follows that, ##\sqrt{8}=2\sqrt{c}+2\sqrt{2c}##

Am i missing something here!

i got it!

##\sqrt{c}##=##\dfrac {\sqrt {8}}{2+2\sqrt{2}}##

⇒##c=0.34314575## Bingo guys! Phew i took time on this men!

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