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 circles x2+y2+2a′x+2b′y+c′=0{x^2} + {y^2} + 2a'x + 2b'y + c' = 0 and 2x2+2y2+2ax+2by+c=02{x^2} + 2{y^2} + 2ax + 2by + c = 0 intersect orthogonally, then
View answer and explanation
Correct option: C
aa′+bb′=c2+c′aa' + bb' = \frac{c}{2} + c'
Option C (aa′+bb′=c2+c′aa' + bb' = \frac{c}{2} + c') is correct because the condition for orthogonal intersection of two circles (2(g1g2+f1f2)=c1+c22(g_1g_2 + f_1f_2) = c_1 + c_2) is applied after adjusting the second equation.

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.