Skip to main content
Back to CIRCLES

JEE MATHEMATICS MCQ: CIRCLES

1 Easy
1 of 1
Application +4 | -1
The equation to the chord of the circle x2+y2=9{x^2} + {y^2} = 9 whose middle point is (1,−2)\left( {1, - 2} \right) is
View answer and explanation
Correct option: C
x−2y−5=0x - 2y - 5 = 0
Option D (x−2y+5=0x - 2y + 5 = 0) is correct because the equation of the chord with a given midpoint (1, -2) is found using the formula T = S', where T is the equation of the tangent-like expression and S' is the value of the circle equation at the midpoint.

More Questions from 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.