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
77 views
in Differential Equations by (94.8k points)
closed by
The differential equation of all circles passing through the origin and having their centres on the x-axis is (1) `x^2=""y^2+""x y(dy)/(dx)` (2) `x^2=""y^2+"3"x y(dy)/(dx)` (3) `y^2=x^2""+"2"x y(dy)/(dx)` (4) `y^2=x^2""-"2"x y(dy)/(dx)`
A. `y^(2)=x^(2)+2xy(dy)/(dx)`
B. `y^(2)=x^(2)-2xy(dy)/(dx)`
C. `x^(2)=y^(2)+xy(dy)/(dx)`
D. `x^(2)=y^(2)+3xy(dy)/(dx)`

1 Answer

0 votes
by (95.5k points)
selected by
 
Best answer
Correct Answer - A
The equation of the family of circles passing through the origin and having their centres on x-axis is
`(x-a)^(2)+(y-0)^(2)=a^(2)or, x^(2)+y^(2)-2ax=0" …(i)"`
Differentiating w.r.to x, we get
`2x+2y(dy)/(dx)-2a=0rArr a= x+y(dy)/(dx)`
Substituting the value of a in (i), we get
`x^(2)+y^(2)-2x^(2)-2xy(dy)/(dx)=0 or, y^(2)=x^(2)+2xy(dy)/(dx)`
as the required differential equation.

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

...