<s>[INST] From any arbitrary point $P$ on $y =\cos x$ tangents $PA$ and $PB$ are drawn to a circle which passes through From any arbitrary point $P$ on $y =\cos x$ tangents $PA$ and $PB$ are drawn to a circle which passes  through the points $(1,0)$ and $(3,0)$ and touches the circle $x^2+y^2-2x-8=0$ and have its centre in first quadrant. Find locus of circumcentre of $\triangle PAB$
My approach : 
The given circle is having centre at $(1,0)$ and radius $3$. Please suggest how to proceed in this problem. thanks.
 [/INST] Given Circle
The fixed circle described in your problem is actually 
$$(x-2)^2+(y-\frac{\sqrt{5}}{2})^2=\frac{9}{4}$$
This is because its tangent circle is $(x-1)^2+y^2=9$ is centered in $(1,0)$ which lies on the given circle, which means that its diameter is equal to the radius of the tangent circle, so it is $\frac{3}{2}$. 
The $x$ coordinate $a$ of the center is 2 because $(1,0)$ and $(3,0)$ lie on the circle.
The $y$ coordinate $b$ of the center can be found by plugging $(1,0)$ into the equation $(x-2)^2+(y-b)^2=\frac{9}{4}$.
Circumcenter
The circumcenter of $\triangle PAB$ is the midpoint of $PO$ where $O=(a,b)$ is the center of the given circle. This is because $\angle PAO=\frac{\pi}{2}$.
Locus
For the point $P=(x,y=cos(x))$, the circumcenter is $C=(\frac{x+a}{2},\frac{y+b}{2})$ so its locus is a modified cosine line:
$$ Y = \frac{\cos(2X-a)+b}{2}$$
where $a=2$ and $b=\frac{\sqrt{5}}{2}$.
</s>