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.
On
Here is a more analytical approach. Let $S_1 : x^2+y^2-2x-8=0$. Let $S_2$ be a member of the set of circles passing through $(1,0)$ and $(3,0)$. So, $S_2 : (x-1)(x-3)+(y-0)(y-0) + \lambda y = 0$. Now, equation of radical axis is $S_1 - S_2 =0$, or, $L: 2x - \lambda y - 11 =0$. For the two circles to touch each other, perpendicular distance of centre from radical axis must be equal to the radius. Therefore, perpendicular distance of centre of $S_1$ i.e $(1,0)$ from $L$ is equal to $r = 3$. $$\frac{|2+0-11|}{\sqrt{4+\lambda^2}} = 3 \implies \lambda = ±\sqrt{5}$$. Circumcentre of $\triangle PAB$ is the midpoint of $P$ and centre of $S_2$. Solve the rest.
Given Circle
The fixed circle described in your problem is actually $$(x-2)^2+\left(y-\frac{\sqrt{5}}{2}\right)^2=\frac{9}{4}$$ 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 $b=\frac{\sqrt{5}}{2}$.