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 = $f\left(x\right)$ represented by the locus of the centre of the circumcircle of triangle $PAB$, then
Range of $y=f\left(x\right)$ is

• A. $[-2,1]$
• B. $[-1,4]$
• C. $[0,2]$
• D. $[2,3]$

Answer 1

The correct option is D $[2,3]$


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
$h=\frac{t+4}{2}$ and $k=\frac{\sin t-5}{2}$
Eliminating $t$, we have
$k=\frac{\sin (2h-4)+5}{2}$
or $y=\frac{\sin (2x-4)+5}{2}$

Thus, the range of
$y=\frac{\sin (2x-4)+5}{2}$
is $[2,3]$