Consider the circle x 2 + y 2 – 10x – 6y + 30 = 0 . Let O be the centre of the circle and…
Mathematics · JEE Advanced · NTA Exams — Co-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
- area of quadrilateral OACB = 4
- the radical axis for the family of circles S=0 is x+y=10
- the smallest possible circle of the family S = 0 is x2 + y2 – 12x – 4y + 38 = 0
- 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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