Find the segment CR in the figure below?

Question

In the figure below, MD=2, CD=16 is RP=5.calculate CR.($S:CR=\sqrt3)$

The figure above is not to scale.

I try: Extending PR to A on the circumference we will have $\triangle ARB_{(right)}$

$AR=16 \implies AP = 16 - 5=11$

$ \triangle CPB: PR^2 = CR.BR$

$OA=R \implies (2R)^2=16^2+BR^2$

$O1=r \implies 4R^2 = 256 + (2r-CR)^2$

Answer 1

This doesn't follow the method started by the OP, so I'm not sure how useful it is, but a simpler method is to use the theorem that if $X$ is a point inside or outside a circle, and two lines $L_1$, $L_2$ through $X$ intersect the circle at $A, B$ and $C, D$ respectively, then $|XA||XB|=|XC||XD|$. Using this on the smaller circle with $X = R$ gives $|RC||RB|=|RP|^2=25$ (because a reflection of $P$ in line $CB$ will lie on the circle at the same distance from $R$ as $P$) and on the larger circle with $X = C$ gives $|RC|(|RB|+|RC|)=|CN||CM|=2 \times 14 =28$, so $|RC|^2 = 3$.

Answer 2

You've done good work, even including doing some that's not used with the method used here. Below is your diagram with a few lines and points added:

As shown, $OM$ is drawn and $N$ is added where $MC$ crosses the circle. Also, using that $MN$ and $AR$ are parallel chords, then from $O$ is drawn the perpendicular bisector of $AR$ (and $MN$), crossing $AR$ at $T$ and $MN$ at $S$. To avoid confusion with the point $R$, have $\lvert OA\rvert = \lvert OM\rvert = r$ instead, so your relation then becomes

$$(2r)^2 = 16^2 + \lvert BR\rvert^2 \;\;\to\;\; \lvert BR\rvert = 2\sqrt{r^2 - 64} \tag{1}\label{eq1A}$$

Using the Pythagorean theorem with $\triangle OAT$, and $\lvert AT\rvert = \frac{1}{2}\lvert AR\rvert = 8$, gives that

$$\lvert OT\rvert^2 + \lvert AT\rvert^2 = \lvert OA\rvert^2 \;\;\to\;\; \lvert OT\rvert = \sqrt{r^2 - 64} \tag{2}\label{eq2A}$$

With $\triangle OMS$, using $\lvert DS\rvert = \lvert AT\rvert = 8 \;\;\to\;\; \lvert MS\rvert = 6$, the Pythagorean theorem here gives

$$\lvert OS\rvert^2 + \lvert MS\rvert^2 = \lvert OM\rvert^2 \;\;\to\;\; \lvert OS\rvert = \sqrt{r^2 - 36} \tag{3}\label{eq3A}$$

Next, we have

$$\lvert CR\rvert = \lvert OS\rvert - \lvert OT\rvert = \sqrt{r^2 - 36} - \sqrt{r^2 - 64} \tag{4}\label{eq4A}$$

Using \eqref{eq1A} and \eqref{eq4A} with your relation for $\triangle CPB$, we get

$$\lvert PR\rvert^2 = \lvert CR\rvert\cdot\lvert BR\rvert \;\;\to\;\; 25 = 2\sqrt{r^2 - 64}\left(\sqrt{r^2 - 36} - \sqrt{r^2 - 64}\right) \tag{5}\label{eq5A}$$

For simpler algebra, let

$$x = r^2 - 64 \;\;\to\;\; r^2 - 36 = x + 28 \tag{6}\label{eq6A}$$

Thus, \eqref{eq5A} becomes

$$\begin{equation}\begin{aligned} \frac{25}{2} & = \sqrt{x}\left(\sqrt{x+28}-\sqrt{x}\right) \\ x + \frac{25}{2} & = \sqrt{x(x+28)} \\ x^2 + 25x + \frac{625}{4} & = x^2 + 28x \\ x & = \frac{625}{12} \end{aligned}\end{equation}\tag{7}\label{eq7A}$$

Also, $x + 28 = \frac{625 + 28(12)}{12} = \frac{961}{12}$. Using this, \eqref{eq7A} and \eqref{eq6A} in \eqref{eq4A} gives

$$\lvert CR\rvert = \sqrt{\frac{961}{12}} - \sqrt{\frac{625}{12}} = \frac{31}{2\sqrt{3}} - \frac{25}{2\sqrt{3}} = \frac{6}{2\sqrt{3}} = \sqrt{3} \tag{8}\label{eq8A}$$