trig involving surds

[nanase]

hello guys!
I came across this question in surds chapter :

I tried drawing trapezium connecting both centres of circles and the tangent points, but I am clueless after that
please help me out

 

[LCKurtz]

hello guys!
I came across this question in surds chapter :

I tried drawing trapezium connecting both centres of circles and the tangent points, but I am clueless after that
please help me out

I will start with another picture and give you some ideas to get you going.

From triangle ADC [MATH]a^2 + (2-r)^2 = (2+r)^2[/MATH] and from triangle BCE [MATH](1-r)^2 + b^2 =(1+r)^2[/MATH]. From the triangle with AB as its hypotenuse and the red line as its base you get [MATH](a+b)^2 + 1 = 3^2[/MATH]. That's three equations in three unknowns. Simplify the first two and you should see a way to get [MATH]a,~b[/MATH], and [MATH]r[/MATH], which is a bonus of two extra variables. Come back if you are still stuck.

 

[nanase]

I will start with another picture and give you some ideas to get you going.
View attachment 13498
From triangle ADC [MATH]a^2 + (2-r)^2 = (2+r)^2[/MATH] and from triangle BCE [MATH](1-r)^2 + b^2 =(1+r)^2[/MATH]. From the triangle with AB as its hypotenuse and the red line as its base you get [MATH](a+b)^2 + 1 = 3^2[/MATH]. That's three equations in three unknowns. Simplify the first two and you should see a way to get [MATH]a,~b[/MATH], and [MATH]r[/MATH], which is a bonus of two extra variables. Come back if you are still stuck.

sir, thank you very much!
I was able to see from your diagram and the coloured line.
from the first and second equation I can get a and b in terms of r. Then I just substituted them into the third equation
I obtained 6 - 4 [MATH]sqroot 2[/MATH] as the answer.
Thank you for your help sir, appreciate it very much since nobody around can help me here.