Let a circle C₁ equiv x^2 + y^2 - 4x + 6y + 1 = 0 and circle C₂ is such that it's

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10-1.Circle and System of Circles

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Let a circle $C_1 \equiv x^2 + y^2 - 4x + 6y + 1 = 0$ and circle $C_2$ is such that it's centre is image of centre of $C_1$ about $x-$axis and radius of $C_2$ is equal to radius of $C_1$, then area of $C_1$ which is not common with $C_2$ is -

A

$10\pi + 3\sqrt 3$

B

$10\pi$

C

$8\pi - 6\sqrt 3$

D

$8\pi + 6 \sqrt 3$

Solution

$x$ – intercept = radius $ = 2\sqrt 3 $

$\therefore $ Required Area $ = 10\pi + \left( {\frac{{\sqrt 3 }}{4}.12 – 2\pi + \frac{{\sqrt 3 }}{4}.12} \right)$

$ = 8\pi + 6\sqrt 3 $

Standard 11

Mathematics

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