Skip to main content
Back to CIRCLES

JEE MATHEMATICS MCQ: CIRCLES

1 Easy
1 of 1
Application +4 | -1
The set of values of aa for which the point (a−1,a+1)(a-1, a+1) lies outside the circle x2+y2=8x^2+y^2=8 and inside the circle x2+y2−12x+12y−62=0x^2+y^2-12 x+12 y-62=0, is
View answer and explanation
Correct option: C
(−32,−3)∪(3,32)(-3 \sqrt{2},-\sqrt{3}) \cup(\sqrt{3}, 3 \sqrt{2})
Option C ((−32,−3)∪(3,32))((-3 \sqrt{2},-\sqrt{3}) \cup(\sqrt{3}, 3 \sqrt{2})) is correct because it represents the set of 'a' values that satisfy both conditions: the point lies outside the first circle and inside the second circle.

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.