JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
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 \(4\)
Answer & Solution
Correct Answer
(D) \(4\)
Step-by-step Solution
Detailed explanation
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…
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