10-1.Circle and System of Circles
hard

Let $C_1$ and $C_2$ be the centres of the circles $x^2 + y^2 -2x -2y -2 = 0$ and $x^2 + y^2 - 6x-6y + 14 = 0$ respectively. If $P$ and $Q$ are the points of intersection of these circles, then the area (in sq. units) of the quadrilateral $PC_1QC_2$ is ............. $\mathrm{sq. \, units}$

A

$8$

B

$6$

C

$9$

D

$4$

(JEE MAIN-2019)

Solution

Since circles are orthogonal and have equal radii therefore the quadirlateral $P{Q_1}Q{C_2}$ is aspuare.

Hence area $ = 2 \times 2 = 4$sp. units.

Since circles are orthogonal and have equal radii therefore the quadrilateral $P{C_1}Q{C_2}$ is a spuare.

Hence area $ = 2 \times 2 = 4$ sq. units.

Standard 11
Mathematics

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