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
778 views
in Circles by (82.5k points)
closed by
A straight line moves such that the algebraic sum of the perpendiculars drawn to it from two fixed points is equal to `2k` . Then, then straight line always touches a fixed circle of radius. `2k` (b) `k/2` (c) `k` (d) none of these
A. 2k
B. k/2
C. k
D. none of these

1 Answer

0 votes
by (88.1k points)
selected by
 
Best answer
Correct Answer - C
Let the fixed points be `A(a, 0) and B(-a, 0)` and let the straight line be` y=mx+c`. Then,
`(mx+c)/(sqrt(1+m^(2)))+(-mx+c)/(sqrt(1+m^(2)))=2k " " ` [Given]
`rArr c=ksqrt(1+m^(2))`
Thus, the straight line is `y=mx+ksqrt(1+m^(2))`. Clearly, it touches the circle `x^(2)+y^(2)=k^(2)` whose radius is k.

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

...