Question bank/JEE Main/Mathematics

JEE MainMathematicsToughCircles

Let PQ be the common tangent to circles x^2 + y^2 + 2x = 0 and
x^2 + y^2 - 6x = 0 as shown in figure, then find the area of the
shaded region PQR.
ekuk fn, x, o`Ùkksa x^2 + y^2 + 2x = 0 vkSj x^2 + y^2 - 6x = 0

Show answer and solution
Let centres be C_1 and C_2
Length of DCT
(PQ)=(C1C2)2−(r1−r2)2=16−1=23= \sqrt{\left( C_{1}C_{2} \right)^{2} - \left( r_{1} - r_{2} \right)^{2}} = \sqrt{16 - 1} = 2\sqrt{3}
Area of quadrilateral
PQC2C1=12×(1+3)×23=43{PQC}_{2}C_{1} = \frac{1}{2} \times (1 + 3) \times 2\sqrt{3} = 4\sqrt{3}
C1M=23C_{1}M = 2\sqrt{3}
tanθ=223⇒θ=π6tan\theta = \frac{2}{2\sqrt{3}} \Rightarrow \theta = \frac{\pi}{6}
∴∠PC1R=2π3&∠QC2R=π3\therefore\angle{PC}_{1}R = \frac{2\pi}{3}\&\angle{QC}_{2}R = \frac{\pi}{3}
Area of Circular sector
RC1P=12×2π3=π3{RC}_{1}P = \frac{1}{2} \times \frac{2\pi}{3} = \frac{\pi}{3}
Area of Circular sector
RC2Q=12×π3×9=3π2{RC}_{2}Q = \frac{1}{2} \times \frac{\pi}{3} \times 9 = \frac{3\pi}{2}
∴ Required area
=43−π3−3π2=43−11π6= 4\sqrt{3} - \frac{\pi}{3} - \frac{3\pi}{2} = 4\sqrt{3} - \frac{11\pi}{6}

Source: Grass-G / M-26 Mathematics Paper 1 15 Jan.docx

More JEE Main questionsChapter practice

Need a video for this idea?

Scan any doubt from your notes and get a whiteboard explanation.

Scan a doubtView plans