JEE Main 2022 — Mathematics Coordinate Geometry

Given,

$${x}^{2}+{y}^{2}-x+2y=\frac{11}{4}$$x2+y2−x+2y=411​

On rearranging terms we get,

$$\Rightarrow {(x-\frac{1}{2})}^{2}+{(y+1)}^{2}={(2)}^{2}$$⇒(x−21​)2+(y+1)2=(2)2

Now given in $\Delta PQR$ΔPQR, $\angle PCR=\frac{\pi }{4}$∠PCR=4π​ so $\angle PQR=22\frac{1}{2}$∠PQR=2221​ by circle property, radius is $2$2 so $RC=QC=2$RC=QC=2

Now, $PR=QR\mathrm{sin}22\frac{1}{2}$PR=QRsin2221​$=4\mathrm{sin}\frac{\pi }{8}$=4sin8π​

And $PQ=QR\mathrm{cos}22\frac{1}{2}=4\mathrm{cos}22\frac{1}{2}$PQ=QRcos2221​=4cos2221​

Now area of $\Delta PQR=\frac{1}{2}PR\times PQ$ΔPQR=21​PR×PQ

$$=\frac{1}{2}(4\mathrm{sin}\frac{\pi }{8})(4\mathrm{cos}\frac{\pi }{8})$$=21​(4sin8π​)(4cos8π​)

$$=4\times (2\mathrm{sin}\frac{\pi }{8}\mathrm{cos}\frac{\pi }{8})$$=4×(2sin8π​cos8π​)

$$=4\mathrm{sin}\frac{\pi }{4}=\frac{4}{\sqrt{2}}=2\sqrt{2}$$=4sin4π​=2​4​=22​