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The circle \mathrm{x^2+y^2=1} cuts the \mathrm{x} -axis at \mathrm{P \: and \: Q}, another circle with center at \mathrm{\: Q} and variable radius intersects the first circle at \mathrm{R} above the \mathrm{\mathrm{x}}-axis and the line segment \mathrm{PQ} at \mathrm{S}. Find the maximum area of the triangle \mathrm{QSR.}
 

Option: 1

\frac{\sqrt3}{9}


Option: 2

\frac{4\sqrt3}{9}


Option: 3

\frac{2\sqrt3}{9}


Option: 4

3\sqrt3


Answers (1)

best_answer

The given circle.
\mathrm{x^2+y^2=1 \text { is }}      .........(i)
with center at \mathrm{O}(0,0) and radius 1 . It cuts \mathrm{x}-axis at the points when \mathrm{y}=0 then \mathrm{x}= \pm 1 i.e. at \mathrm{P(-1,0) \: and \: Q(1,0)}

Equation of circle with center at \mathrm{Q(1,0)} and radius \mathrm{r} is

(\mathrm{x}-1)^2+(\mathrm{y}-0)^2=\mathrm{r}^2(0<\mathrm{r}<2)..........(ii)

Solving (i) and (ii), we get

\mathrm{x=\frac{2-r^2}{2} \text { and } y= \pm \frac{r \sqrt{\left(4-r^2\right)}}{2}}

but \mathrm{\mathrm{R}\: above \: the \, \mathrm{x}-axis.}

\mathrm{\therefore \quad R \equiv\left(\frac{2-r^2}{2}, \frac{r \sqrt{4-r^2}}{2}\right)}

\mathrm{ \text { so, } S Q=r \text { and } M R=\frac{r \sqrt{4-r^2}}{2} }
\mathrm{ \therefore \quad \text { Area of } \Delta Q S R=\frac{1}{2} Q S M R }
\mathrm{ \Delta=\frac{1}{2} r \cdot \frac{r \sqrt{4-r^2}}{2} }
\mathrm{ \Delta(\text { say })=\frac{r^2 \sqrt{\left(4-r^2\right)}}{4} }
\mathrm{ \qquad \Delta^2=\frac{r^4}{16}\left(4-r^2\right)=A(\text { say }) }
\mathrm{ \therefore \quad A=\frac{1}{16}\left(4 r^4-r^6\right) }
\mathrm{ \therefore \quad \frac{\mathrm{dA}}{\mathrm{dr}}=\frac{1}{16}\left(16 \mathrm{r}^3-6 \mathrm{r}^5\right) }
\mathrm{ \text { and } \quad \frac{\mathrm{d}^2 \mathrm{~A}}{\mathrm{dr}^2}=\frac{1}{16}\left(48 \mathrm{r}^2-30 \mathrm{r}^4\right) }
\text { for maximum and minimum area, } \frac{\mathrm{dA}}{\mathrm{dr}}=0

we get \mathrm{r^2=\frac{8}{3}}
\therefore \quad\left(\frac{\mathrm{d}^2 \mathrm{~A}}{\mathrm{dr}^2}\right)_{\mathrm{r}-\sqrt{\frac{8}{3}}}=\frac{1}{16}\left(48 \times \frac{8}{3}-30 \times \frac{64}{9}\right)<0
\therefore \mathrm{A} is maximum. Hence \Delta is also maximum.
\therefore \text { Maximum value of } \begin{aligned} \Delta & =\frac{1}{2} \times \frac{8}{3} \times \frac{1}{2} \sqrt{\left(4-\frac{8}{3}\right)} \\ \end{aligned}

                                                \mathrm{ =\frac{2}{3} \times \frac{2}{\sqrt{3}} }

                                                \mathrm{=\frac{4 \sqrt{3}}{9} \text { sq. units. } }

Hence option 2 is correct.

Posted by

HARSH KANKARIA

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