Going round in circles

Let $\angle CAD=x$ . Then considering circle $C_3$ where $AD=CD=r_{C_3},$ , we see that $\angle ACD=\angle CAD=x$ . It then follows that $\angle BDC=\angle ACD+\angle CAD=x+x=(1+1)x=2x$ . Similarly, considering circle $C_2$ where $BC=BD$ , we have $\angle BCD=\angle BDC=2x.$

Next, we also deduce that $\angle BCA=\angle BCD+\angle ACD=2x+x=(2+1)x=3x$ . Considering circle $C_1$ where $AB=AC$ , we have $\angle ABC=\angle BCA=3x$ . To solve for $x$ :

$\angle CBD+\angle BCD+\angle BDC=\pi$

$3x+2x+2x=\pi$

$(3+4)x=\pi$

$7x=\pi$

$x=\frac{\pi}{7}$

Alternatively,

$\angle ABC+\angle ACB+\angle BAC=\pi$

$3x+3x+x=\pi$

$(6+1)x=\pi$

$7x=\pi$

$x=\frac{\pi}{7}$

Yet another method is:

$\angle ADC=\angle CBD+\angle BCD=3x+2x=(3+2)x=5x$

$\angle ADC+\angle ACD+\angle CAD=\pi$

$5x+x+x=\pi$

$(5+2)x=\pi$

$7x=\pi$

$x=\frac{\pi}{7}$

Therefore, $\frac{X}{Y}=\frac{2\pi 2r{C_1}}{\frac{\pi}{7} 2r_{C_1}}=\frac{2}{\frac{1}{7}}=2 \times 7=\boxed{14}$