In acute triangle $\triangle ABC$, point $R$ lies on the perpendicular bisector of $AC$ such that $\overline{CA}$ bisects $\angle BAR$. Let $Q$ be the intersection of lines $AC$ and $BR$. The circumcircle of $\triangle ARC$ intersects segment $\overline{AB}$ at $P\neq A$, with $AP=1$, $PB=5$, and $AQ=2$. Compute $AR$.
1. **Identify the given information and setup the problem:**
   - $\triangle ABC$ is an acute triangle.
   - Point $R$ lies on the perpendicular bisector of $AC$.
   - $\overline{CA}$ bisects $\angle BAR$.
   - $Q$ is the intersection of lines $AC$ and $BR$.
   - The circumcircle of $\triangle ARC$ intersects segment $\overline{AB}$ at $P \neq A$.
   - Given: $AP = 1$, $PB = 5$, and $AQ = 2$.

2. **Analyze the cyclic quadrilateral $APCR$:**
   - Since $APCR$ is a cyclic quadrilateral, we can use properties of cyclic quadrilaterals.
   - In cyclic quadrilateral $APCR$, $\angle APR = \angle ACR$ and $\angle PAR = \angle PCR$.

3. **Use the given lengths to find relationships:**
   - Given $AP = 1$ and $PB = 5$, we have $AB = AP + PB = 1 + 5 = 6$.
   - Given $AQ = 2$.

4. **Use the fact that $R$ lies on the perpendicular bisector of $AC$:**
   - Since $R$ is on the perpendicular bisector of $AC$, $AR = CR$.

5. **Use the angle bisector theorem:**
   - Since $\overline{CA}$ bisects $\angle BAR$, by the angle bisector theorem, $\frac{AQ}{QC} = \frac{AB}{BR}$.
   - Let $QC = x$. Then $\frac{2}{x} = \frac{6}{BR}$, so $BR = 3x$.

6. **Use Ptolemy's theorem in cyclic quadrilateral $APCR$:**
   - Ptolemy's theorem states that in a cyclic quadrilateral, the sum of the products of its two pairs of opposite sides is equal to the product of its diagonals.
   - Applying Ptolemy's theorem to $APCR$: $AP \cdot CR + AR \cdot PC = AC \cdot PR$.
   - Since $AR = CR$ and $AP = 1$, we have $1 \cdot AR + AR \cdot PC = AC \cdot PR$.

7. **Express $PC$ in terms of $AR$:**
   - Since $P$ lies on $AB$, $PC = PB = 5$.

8. **Solve for $AR$:**
   - Substitute $PC = 5$ into the equation: $1 \cdot AR + AR \cdot 5 = AC \cdot PR$.
   - Simplify: $AR + 5AR = AC \cdot PR$.
   - $6AR = AC \cdot PR$.

9. **Find $AC$ and $PR$:**
   - Since $Q$ is the intersection of $AC$ and $BR$, and $AQ = 2$, $QC = x$.
   - $AC = AQ + QC = 2 + x$.
   - $PR = AR$ (since $R$ is on the perpendicular bisector of $AC$).

10. **Substitute $AC$ and $PR$ into the equation:**
    - $6AR = (2 + x) \cdot AR$.
    - Simplify: $6 = 2 + x$.
    - Solve for $x$: $x = 4$.

11. **Find $AR$:**
    - Since $AR = CR$ and $R$ is on the perpendicular bisector of $AC$, $AR = CR$.
    - $AR = 6$.

The final answer is $\boxed{6}$.