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Let C₁ and C₂ 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 PC₁ QC₂ is

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answer is D.

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Detailed Solution

$\begin{array}{l} 2 g_{1} g_{2} + 2 f_{1} f_{2} = 2 (- 1) (- 3) + 2 (- 1) (- 3) = 12 \\ c_{1} + c_{2} = 14 - 2 = 12 \end{array}$

Since, $2 g_{1} g_{2} + 2 f_{1} f_{2} = c_{1} + c_{2}$

Hence, circles intersect orthogonally

$∴$ Area of the quadrilateral PC₁QC₁

$= 2 \times \frac{1}{2} r_{1} r_{2} = (2) (2) = 4 sq . units$

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