Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
299 views
in Circles by (82.5k points)
closed by
The limiting points of the coaxial system containing the two circles `x^(2)+y^(2)+2x-2y+2=0` and `25(x^(2)+y^(2))-10x-80y+65=0` are
A. (1, -1), (-5, -40)
B. (1, -1), (-1/5, -8/5)
C. (-1, 1) , (1/5, 8/5)
D. (-1, 1), (-1/5, -8/5)

1 Answer

0 votes
by (88.1k points)
selected by
 
Best answer
Correct Answer - C
The equation of the radical axis of the given circles is `4x+2y-1=0`. Therefore, the equation of family of coaxial circles is
`x^(2)+y^(2)+2x-2y+2+lambda(4x+2y-1)=0`
`rArr x^(2)+y^(2)+2x(1++2lambda)+2y(lambda-1)+2-lambda=0`
The coordinates of the centre and radius of this circle are
`-(2lambda+1), -(lambda-1) and sqrt((2lambda+1)^(2)+(lambda-1)^(2)-2)`
For limiting points, we must have
Radius=0
`rArr (2lambda+1)^(2)+(lambda-1)^(2)+lambda-2=0 rArr lambda=0, -3//5`
Hence, the limiting points are (-1, 1) and (1/5, 8/5)

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...