Question

Tangents $PA$ and $PB$ are drawn to the circle $(x-4{)}^2+$ $(y-5{)}^2=4$ from the point $P$ on the curve $y=\sin x$ where $A$ and $B$ lie on the circle. Consider the function $y=f(x)$ represented by the locus of the centre of the circumcircle of triangle $PAB$. Then answer the following questions.
The period of $y=f(x)$ is

• A. $2\pi$
• B. $3\pi$
• C. $\pi$
• D. not defined

Answer 1

Answer: (C)

The center of the given circle is $C(4,5)$. Points $P,A,C$ and $B$ are concyclic such that $PC$ is the diameter of the circle. Hence the center D of the circumcircle of $\Delta ABC$ is the midpoint of $PC$.
Then, we have
$h=\frac{t+4}{2}\text{ and }k=\frac{\sin t-5}{2}$
Eliminating t, we have
$k=\frac{\sin (2h-4)+5}{2}$
or $y=\frac{\sin (2x-4)+5}{2}$
or $f^{-1}(x)=\frac{{\sin }^{-1}(2x-5)+4}{2}$
Thus, the range of
$y=\frac{\sin (2x-4)+5}{2}$
is $[2,3]$ and the period is $\pi$.
Also, $f(x)=4$, i.e. $\sin (2x-4)=3$ which has no real solutions.
For $f(x)=1,\sin (2x-4)=-3$ which has no real solutions.
But range of $y=\frac{{\sin }^{-1}(2x-5)+4}{2}$ is $\left[-\frac{\pi }{4}+2,\frac{\pi }{4}+2\right]$