Skip to main content
Back to CIRCLES

JEE MATHEMATICS MCQ: CIRCLES

1 Easy
1 of 1
Analyzing +4 | -1
The shortest distance between the parabola y2=4x{y^2} = 4x and the circle x2+y2+6x−12y+20=0{x^2} + {y^2} + 6x - 12y + 20 = 0 is
View answer and explanation
Correct option: A
42−54\sqrt 2 - 5
Option A (42−54\sqrt 2 - 5) is correct because it represents the shortest distance between the parabola and the circle, calculated along the common normal.

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.