Question

The circle $x^2+y^2=1$ cuts the x-axis at P and Q. Another circle with centre at Q and variable radius intersects the first circle at R above the x-axis and the line segment PQ at S. If A is the maximum area of the triangle QSR then $3\sqrt{3}$ A is equal to _____ .

Answer 1

$x^2+y^2=1\therefore P(1,0),Q(-1,0)$

Also ${(x+1)}^2+y^2=r^2$ is the equation of second circle centered at $Q$ and of radius $r$.

Solving these, we get

$x^2+y^2+2x+1=r^2\Rightarrow 1+2x+1=r^2\Rightarrow x+1=\frac{r^2}{2}$

Putting for $(x+1)$, we get

$y^2=r^2-{(x+1)}^2=r^2-\frac{r^4}{4}$ ....(1)

Again it meets $PQ$ i.e. x-axis or $Y=0$ at $S(-1\pm r,0)$

$\therefore OS=\pm r$ ...(2)

If $\delta$ be the are of triangle $QSR$, then

$\Delta =\frac{1}{2}$ base $\times$ height

From (1) and (2) ${\Delta }^2=\frac{1}{4}{r}^2.y^2=\frac{1}{4}(r^2-\frac{r^4}{4})$

$\Rightarrow Z={\Delta }^2=\frac{1}{4}(r^4-\frac{r^6}{4})$ ...(3)

$\therefore \frac{dZ}{dr}=\frac{1}{4}(4r^3-\frac{3r^5}{2})=0\Rightarrow r^2=\frac{8}{3}$

$\therefore \frac{d^{2}Z}{d{r}^2}=\frac{1}{4}(12r^2-\frac{15r^4}{2})=\frac{3}{4}.\frac{8}{3}(4-\frac{5}{2}.\frac{8}{3})$

$\therefore \delta$ is maximum when $r^2=\frac{8}{3}$

Also $y^2=r^2(1-\frac{r^2}{4})=\frac{8}{3}(1-\frac{1}{4}.\frac{8}{3})=\frac{8}{3}.\frac{1}{3}$

$\therefore y=\pm \frac{2\sqrt{2}}{3}$

or $y=\frac{2\sqrt{2}}{3}$ as $R$ is above $PQ$ i.e. $x$ is

Putting for $r^2$ in (3), we get

${\Delta }^2=\frac{1}{4}(\frac{64}{9}-\frac{1}{4}.\frac{512}{27})=\frac{16}{27}$

$\Rightarrow \Delta =\frac{4}{3\sqrt{3}}\therefore A=4$