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 center of the circumcircle of triangle PAB. Then answer the following questions.
The range of `y=f(x)` is

Options

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

Answer

4

Explanation

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(h,k) of the circumcircle of `Delta ABC` is the midpoint of PC.
Then, we have
`h=(t+4)/(2)` and `k =(sin t +5)/(2)`
Eliminating t, we have
`k=(sin (2h-4)+5)/(2)`
or `y=f(x)=(sin (2x-4)+5)/(2)`
`:. f^(-1)(x) = (sin^(-1)(2x-5)+4)/(2)`
Thus, the range of
`y=(sin (2x-4)+5)/(2)`
is `[2,3]` and the priod is `pi`.
Al,so `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= (sin ^(-1)(2x-5)+4)/(2) ` is `[-(pi)/(3) +2,(pi)/(4)+2]`