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.

  • $\begingroup$ The circle in "your approach" does not pass through the points (1,0) and (3,0). $\endgroup$ – sds Jun 16 '14 at 15:46

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 $(x-2)^2+(y-b)^2=\frac{9}{4}$.


The circumcenter of $\triangle PAB$ is the midpoint of $PO$ where $O=(a,b)$ is the center of the given circle. This is because $\angle PAO=\frac{\pi}{2}$.


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 $b=\frac{\sqrt{5}}{2}$.

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  • $\begingroup$ The only modification I'd make is that the tangents do not exist when $P$ is inside the circle $(x-2)^2+(y-\frac{\sqrt{5}}{2})^2=\frac94$, so the locus $y=\frac{\cos{(2x-2)}+\frac{\sqrt{5}}{2}}{2}$ is not defined for $1.26<x<1.99$. (+1) $\endgroup$ – Jan-Magnus Økland Oct 10 '19 at 6:28

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