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Distance Formula in Cartesian Plane

Let C1 and C2...

Question

Let $C_{1}$ and $C_{2}$ be the centres of the circles $x^{2} + y^{2} - 2 x - 2 y - 2 = 0$ and $x^{2} + y^{2} - 6 x - 6 y + 14 = 0$ respectively. If $P$ and $Q$ are the points of intersection of these circles, then the area (in sq. units) of the quadrilateral $P C_{1} Q C_{2}$ is :

4

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6

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8

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Solution

The correct option is A $4$
Given circle are
$x^{2} + y^{2} - 2 x - 2 y - 2 = 0$ and $x^{2} + y^{2} - 6 x - 6 y + 14 = 0$
$C_{1} = (1, 1), r_{1} = 2 C_{2} = (3, 3), r_{2} = 2$

As $P$ lies on the circle, so
$P C_{1} = 2, P C_{2} = 2 C_{1} C_{2} = \sqrt{4 + 4} = 2 \sqrt{2} \Rightarrow (P C_{1})^{2} + (P C_{2})^{2} = (C_{1} C_{2})^{2}$

$∴ △ C_{1} P C_{2}$ is a right angle triangle

So, the area of quadrilateral $P C_{1} Q C_{2}$
$= 2 (\frac{1}{2} \times P C_{1} \times P C_{2}) = 2 (\frac{1}{2} \times 2 \times 2) = 4 sq. units$

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