2
$\begingroup$

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.

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.