Chord or Tangent???

From the above figure , $\begin{aligned} sin(\angle DCI) &= \left(\dfrac{1}{2} \right)\\ \angle DCI &= 30^\circ \end{aligned}$

Therefore we can say- Slope of chord $= tan(30^\circ) = 1/\sqrt{3}$

Thus by using point-slope form of a line, the equation of chord becomes:

$\begin{aligned} (y-0) &=(1/\sqrt{3})(x-1)\\ y &= \dfrac{x-1}{\sqrt{3}}\\ \dfrac{x-1}{\sqrt{3}} -y &= 0 \end{aligned}$

Therefore now consider a perpendicular AP is dropped from the centre of the bigger circle i.e $(0,0)$ .

Join point A to any one of the ends of the cord to obtain a right triangle $APQ$

Now usin pythagoras theorem and Distance from a point to a line formula :

$\begin{aligned} AP^2 + PQ^2 &= AQ^2\\ \\ \Rightarrow PQ^2 &=AQ^2 - AP^2\\ \\ \Rightarrow PQ^2 &= 16 - \left( \dfrac{ \left| \dfrac{0-1}{\sqrt{3}} -0 \right| }{ \sqrt { 1+ \frac{1}{3}} } \right)^2\\ \\ \Rightarrow PQ &= \dfrac{\sqrt{63}}{2} \end{aligned}$

Now since we know that when any perpendicular is dropped from the center of a circle to a chord, it divides the chord into two halves.

Therefore: Length of chord $= 2 \cdot PQ = 2 \cdot \dfrac{\sqrt{63}}{2} = \sqrt{63}$

Thus comparing our result with the form given in the question,

$\boxed{x = 63}$

name the centre of cirlce C as O and of circle D as P....

Join PD. Sin $\angle$ DCP = 1/2

therefore $\angle$ DCP = 30' = Slope of the chord = m

as we know y=mx+c. putting value of m and satisfying with the point (1,0)

we get c=- $\frac{1}{\sqrt{3}}$

therefore the equation of chord y= $\frac{1}{\sqrt{3}}$ (x-1) -----------1

the equation of bigger circle is $x^2$ + $y^2$ = 16 ----------2

solving equation 1 and 2 we get x= $\frac {1 \pm 3\ \sqrt{21} }{4}$ & y= $\frac{-3 \pm 3\ \sqrt{21} }{4 \sqrt{3}}$

by distance formula on these four points we get the length of chord as $\sqrt{63}$ So the answer is $\boxed{63}$