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AP EAMCET202421 May 2024Morning ShiftMathematicsCircleActual

2 x-3 y+1=0 and 4 x-5 y-1=0 are the equations of two diameters of the circle S x^2+y^2+2 g x+2 f y-11=0 . Q and R are the points of contact of the tangents drawn from the point P(-2,-2) to this circle. If C is the centre of the circle S=0 , then the area (in square units) of the quadrilateral P Q C R is

Options

  1. A25
  2. B30
  3. C24
  4. D36^

Correct answer

B. 30

Step-by-step solution

Equation fo diameters of the circle are aligned & 2 x-3 y+1=0 ...(i) & 4 x-5 y-1=0 ...(ii) aligned On solving equation (i) and (ii) we get centre if circle S =(-g,-f)=(3,4) g=-3 and f=-4 Radius of circle's, CQ = 9+16+11 =6 aligned & C P= (3+2)^2+(4+2)^2 = 61 & P Q= C P^2-C Q^2 = 61-36 = 25 =5 aligned Now, area of quadrilateral PQCR =2 area of CQP =2 1 2 CQ PQ =6 5=30 sq. units

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