The circumference of the circle x 2 + y 2 – 2x + 8y – q = 0 is bisected by the circle x 2…
Mathematics · JEE Advanced · NTA Exams — Co-ordinate Geometry
The circumference of the circle x2 + y2 – 2x + 8y – q = 0 is bisected by the circle x2 + y2 + 4x + 12y + p = 0, then
p + q is equal to :
p + q is equal to :
- 25
- 100
- 10
- 48
Answer
(C) 10
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