For reference: In the acute triangle $ABC$ of orthocenter $H$ and circumcenter $O$, the height $AP$. If the distance from $O$ to $AC$ is $2$ and $PH = 1$; calculate the power of $P$ in relation the center circumference $O$.(Answer:$ -5$)
My progress:
Points inside the circle have negative power;
$P(P) = d^2 - R^2 = PO^2-AO^2 = r^2 - R^2\\ \triangle AOD: R^2 = 4+AD^2\\ ADE \sim \triangle APC \implies \frac{DE}{CP} = \frac{AD}{AH+1}=\frac{AE}{2AD} \\BH=2OD (property)= 2.2=4\\ \triangle BPH: 4^2 = BP^2+1^2 \therefore BP = \sqrt15\\ \triangle APC \sim \triangle BPH \implies: \frac{PC}{1}=\frac{2AD}{4}=\frac{AH+1}{\sqrt{15}} \implies AD = 2PC\\ \therefore AE = 4DE$
???...
$ - 5$ is not the correct answer.
Note that $BH = 2 OD = 4$ and hence $BP = \sqrt{15}$
We then use the fact that $\triangle ACP \sim \triangle BHP$
$ \displaystyle \implies AC = \frac{4}{\sqrt{15}} (AH+1), ~ PC = \frac{1}{\sqrt{15}} (AH+1)$
If $M$ is the midpoint of $BC$, $ \displaystyle CM = \frac {\sqrt{15} + PC}{2}$
$ \displaystyle PM = CM - PC = \frac {\sqrt{15} - PC}{2} = \frac{14 - AH}{2 \sqrt{15}}$
Power of point $P$ with respect to the circumcenter $O = OP^2 - OA^2$,
$ \displaystyle = OM^2 + PM^2 - (AD^2 + 4)$
$ \displaystyle = \frac{AH^2}{4} + \frac{(14-AH)^2}{60} - \frac{AC^2}{4} - 4$
$ \displaystyle = \frac{AH^2}{4} + \frac{(14-AH)^2}{60} - \frac{4 (AH+1)^2}{15} - 4 = - AH - 1$
So it depends on $AH$ and unless $\triangle ABC$ is isosceles with $\angle A = \angle B$, the answer cannot be $ - 5$.