JEE MainMathematicsModerateCircles

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.

Show answer and solution

Let centres be C_1 and C_2
Length of DCT
(PQ) =

\sqrt{\left( C_{1}C_{2} \right)^{2} - \left( r_{1} - r_{2} \right)^{2}} = \sqrt{16 - 4} = 2\sqrt{3}

Area of quadrilateral

{PQC}_{2}C_{1} = \frac{1}{2} \times (1 + 3) \times 2\sqrt{3} = 4\sqrt{3}

C_{1}M = 2\sqrt{3}

\tan\theta = \frac{2}{2\sqrt{3}} \Rightarrow \theta = \frac{\pi}{6}

\therefore\angle{PC}_{1}R = \frac{2\pi}{3}\&\angle{QC}_{2}R = \frac{\pi}{3}

Area of circular sector RC_1 P =

\frac{1}{2} \times \frac{2\pi}{3} = \frac{\pi}{3}

Area of circular sector RC_2 Q =

\frac{1}{2} \times \frac{\pi}{3} \times 9 = \frac{3\pi}{2}

\therefore

Required area =
4

\sqrt{3} - \frac{\pi}{3} - \frac{3\pi}{2} = 4\sqrt{3} - \frac{11\pi}{6}

Source: Grass-G / M-23 Mathematics Paper 31 Dec FST.docx

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