Consider the circle x 2 + y 2 – 10x – 6y + 30 = 0 . Let O be the centre of the circle and…

Mathematics · JEE Advanced · NTA ExamsCo-ordinate Geometry

Consider the circle x2 + y2 – 10x – 6y + 30 = 0 . Let O be the centre of the circle and tangent at A(7, 3) and B(5, 1) meet at C. Let S = 0 represents family of circles passing through A and B, then
  1. area of quadrilateral OACB = 4
  2. the radical axis for the family of circles S=0 is x+y=10
  3. the smallest possible circle of the family S = 0 is x2 + y2 – 12x – 4y + 38 = 0
  4. the coordinates of point C are (7, 1)

Answer

(A) area of quadrilateral OACB = 4, (C) the smallest possible circle of the family S = 0 is x 2 + y 2 – 12x – 4y + 38 = 0, (D) the coordinates of point C are (7, 1)

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