Skip to main content
Back to SYSTEM OF CIRCLES

JEE MATHEMATICS MCQ: SYSTEM OF CIRCLES

1 Easy
1 of 1
Application +4 | -1
If the circle x2+y2+4x+22y+c=0{x^2} + {y^2} + 4x + 22y + c = 0 bisects the circumference of the circle x2+y2−2x+8y−d=0{x^2} + {y^2} - 2x + 8y - d = 0, then c+dc + d is equal to
View answer and explanation
Correct option: B
50
Option B (50) is correct because when one circle bisects the circumference of another, the common chord passes through the center of the second circle, leading to the equation c+d=50c + d = 50.

More Questions from SYSTEM OF CIRCLES

Practice All Chapter MCQs →
Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.