A circle $S$ whose radius is $1$ unit, touches the X-axis at point $A$. The centre $O$ of $S$ lies in the first quadrant. The tangent from the origin $O$ to the circle touches it at $T$ and a point $P$ lies on it such that the triangle $OAP$ is a right-angled triangle at $A$ and its perimeter is $8$ unit. The length of $QP$ is ___.
Step-by-Step Solution
Key Concept: Similar triangles combined with perimeter constraint and cubic equation solving determine the configuration.
Given $OA = h$ and $PQ = x$, triangles $OAP$ and $QTP$ are similar, yielding $\frac{AP}{PT} = \frac{OA}{QT} = \frac{OP}{PQ}$, which gives $\frac{x+1}{PT} = \frac{h}{1} = \frac{OP}{x}$. From the perimeter condition $OA + AP + OP = 8$ and solving the resulting cubic equation $h^3 - 3h^2 + 4 = 0$, we get $h = 2$. Then $x = \frac{5}{3}$ and $PQ = \frac{5}{3}$.
Correct Answer: 1.66