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JEE Main 2023 · Mathematics · Straight Lines and Circles · medium

The circles x^2+y^2=25 and x^2+y^2-12x+11=0 intersect at A and B and have centres C1 and C2. The area of the quadrilateral C1AC2B is

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The circles x2+y2=25x^2+y^2=25 and x2+y2−12x+11=0x^2+y^2-12x+11=0 intersect at AA and BB and have centres C1C_1 and C2C_2. The area of the quadrilateral C1AC2BC_1AC_2B is
  1. A. 2424
  2. B. 1212
  3. C. 3030
  4. D. 4848
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Answer: (A) 2424

Rewrite the second circle as (x−6)2+y2=25(x-6)^2+y^2=25, with centre C2(6,0)C_2(6,0) and radius 55; the first has centre C1(0,0)C_1(0,0) and radius 55. Subtracting the equations gives the common chord 12x=3612x=36, i.e. x=3x=3, and then y2=25−9y^2=25-9, so A,B=(3,±4)A,B=(3,\pm4) and AB=8AB=8. Since all four sides equal 55, the quadrilateral is a rhombus with diagonals C1C2=6C_1C_2=6 and AB=8AB=8, perpendicular to each other. Its area is 12⋅6⋅8=24\frac12\cdot6\cdot8=24.
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