If $\frac{r^2}{AP^2}$ can stated in the form $\frac{a}{b}$ where $\gcd(a, b) = 1$. Determine the value of $a + 10b.$ Given a circle $ω$ with center $O$ and length of radius $r$. There is
a point $A$, $B$ on the circle and point $C$ outside the circle so
that $BC$ tangent to the circle $ω$. Then $AC$ intersects the circle
$ω$ at $D$. Then $P$ is the midpoint of $AC$ and $AD = BP = BC$. If $\frac{r^2}{AP^2}$ can stated in the form $\frac{a}{b}$ where $gcd$ $(a, b) = 1$. Determine the value of $a + 10b.$

Suppose $AP = y$, $DP = x$
with the power of point theorem, i get :
$BC^2=CD.AC\Longrightarrow (x+y)^2=2y(y-x)\Longrightarrow x^2+4xy-y^2=0\Longrightarrow y=x(2+\sqrt{5})$
i need an equation containing $x$ and $r$,but I didn't find it, if someone can help, that's great!!!
There are the original question, Page 4, Number 5.
The ratio ${\large{\frac{r^2}{AP^2}}}$ can be computed by brute force as outlined below.

Assume $y=1$.

Since the ratio ${\large{\frac{r^2}{AP^2}}}$ is scale invariant, we can assume $y=1$.
Compute $x$.

By power of a point we get $(x+1)^2=(1-x)(2)$, so
$$x=\sqrt{5}-2$$
Compute $\cos\bigl(\angle{BCP}\bigr)$.

By the law of cosines in triangle $BCP$ we get
$$\cos\bigl(\angle{BCP}\bigr)=\frac{1}{2(x+1)}$$
Compute $BD$.

By the law of cosines in triangle $BCD$ we get
$$BD=\sqrt{3-7x}$$
Compute $\cos\bigl(\angle{CBD}\bigr)$.

By the law of cosines in triangle $CBD$ we get
$$\cos\bigl(\angle{CBD}\bigr)=\frac{3(1-x)}{2(1+x)\sqrt{3-7x}}$$
Compute $\cos\bigl(\angle{OBD}\bigr)$.

Since angles $OBD$ and $CBD$ are complementary we get
$$
\cos\bigl(\angle{OBD}\bigr)
=
\sin\bigl(\angle{CBD}\bigr)
=
\sqrt{1-\cos^2\bigl(\angle{CBD}\bigr)}
=
\frac{\sqrt{748-198x}}{44}
$$
Compute $r$.

By the law of cosines in triangle $OBD$ we get
$$r=\frac{22\sqrt{3-7x}}{\sqrt{748-198x}}$$
Compute ${\large{\frac{r^2}{AP^2}}}$.

Since $AP=y=1$ and $x=\sqrt{5}-2$ we get
$$
\frac{r^2}{AP^2}
=
r^2
=
\frac{22(3-7x)}{34-9x}
=
\frac{1138-422\sqrt{5}}{209}
$$

Thus we've computed ${\large{\frac{r^2}{AP^2}}}$, but contrary to the claim of the problem statement, it's not a rational number.
First we have to prove that points $AOB$ are collinear . For that you have to prove that $∠AOP+∠POD+∠DOC+∠BOC=180^o$ .
$\Rightarrow$ Now , as you can see that $ΔABC$ is a Right triangle . So apply pythagoras theorem . $$(x+y)^2+(2r)^2=(2y^2)$$ $$x^2-3y^2+2xy=-4r^2$$
$\rightarrow$ Now , subsitute $y$ with $x$ from $y=x(2+\sqrt5)$ $$4r^2=x^2(22+10\sqrt5)$$
$\rightarrow$ subsitute $x$ with $y$ . you'll get $$\frac{r^2}{AP^2}=\frac{11+5\sqrt5}{18+8\sqrt5}=\frac{a}{b}$$
$\Rightarrow$ so you'll get $hcf(a,b)=1$ . Now you can use the value of $a$ &amp; $b$ . To get your answer .
$\Rightarrow$ I apologize for calculation mistakes if there is one , or any other mistake . Sorry for that !