Question

The circle $x^2+y^2=1$ cuts the $x$-axis at $P$ & $Q$. Another circle with centre at $Q$ and variable radius intersects the first circle at $R$ above the $x$-axis & the line segment $PQ$ at $S$. Let the maximum area of the triangle $QSR$ be $\frac{k}{m\sqrt{n}}$. Find $k+m+n$ ?

Answer 1

The centre of the circle $x^2+y^2=1$ ...(1)
is $(0,0)$ and radius $OP=1=OQ$.
So the co-ordinates of $Q$ are $(1,0)$.
Let the radius of the variable circle be $r$.
Hence, the equation is ${(x-1)}^2+{\left(y\right)}^2=r^2$ ...(2)
Subtracting (2) from (1) we get, $2x-1=1-r^2$
$x=1-\frac{r^2}{2}=OT$ ...(3)
Now, $RT=\sqrt{O{R}^2-O{T}^2}=\sqrt{1-{(1-\frac{r^2}{2})}^2}$ ...(4)
Now, the area of $△OSR$ is,
$A=\frac{1}{2}.QS.RT$ i.e.$A^2=\frac{1}{4}\left(Q{S}^2\right).\left({RT}^2\right)$
$A^2=\frac{1}{4}{r}^2(r^2-\frac{r^4}{4})$ [using (2) and (4)]
$A^2=\frac{1}{16}(4r^4-r^6)$
Thus, $\frac{d\left(A^2\right)}{dr}=\frac{1}{16}(16r^3-6r^5)=0$ (for extremum)
$\Rightarrow r=2\sqrt{\frac{2}{3}}$
Also, $\frac{d^2\left(A^2\right)}{d{r}^2}=\frac{1}{16}(48r^2-30r^4)=-\frac{16}{3}<0$
when $r=2\sqrt{\frac{2}{3}}$
Hence, the area is maximum at $r=2\sqrt{\frac{2}{3}}$ and $A_{max.}=\frac{4}{3\sqrt{3}}$sq.units