$A$ is a fixed point on a circle of radius $1$ unit, $P$ is a variable point on this circle. A circle is formed touching the $CP,CA$ and the $Arc(AP)$ where $C$ is the center of the original circle. What will be the area enclosed by the locus of center of new circle so formed?

My Approach: Shifting the origin and rotating the axis. I consider the origin of the original circle to be at the origin and point A to be fixed at (1,0). Now the point P can be written as $(\cos{\theta} , \sin{\theta})$ The center of circle touching the radii CP and CA should lie on the angle bisector. Hence if $(h,k)$ is the center of this circle and $r$ is its radius then $|k|=r$ and $\sin{\theta} = \frac{r}{1-r}$ and $h = (1-r)\cos{\theta}$ using this I obtain the locus of the curve but I am not sure how to proceed further.

Also there could probably be a much better way to approach this. All hints/explanations/solutions are welcome. Thanks!


1 Answer 1


Let $C=(0,0)$, $A=(0,1)$.

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Due to the symmetry it's enough to consider just the first quadrant, for which $x$-coordinate of the inscribed circle is $r$, and the $y$-coordinate is $y(r)=\sqrt{1-2r}$, so the area is \begin{align} S&=4\int_0^{1/2}\sqrt{1-2r}\, dr =\frac43 . \end{align}

  • $\begingroup$ How does the y coordinate comes out to be $y(r)=\sqrt{1-2r}$ ? $\endgroup$
    – marks_404
    Jun 10, 2021 at 14:21
  • $\begingroup$ @protus: $|CO|=1-r$ is hypotenuse, $r$ is one leg, what is another? $\endgroup$
    – g.kov
    Jun 10, 2021 at 14:24

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