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
456 views
in Differential Equations by (28.2k points)
closed by

Consider the equation of all circles which pass through the origin and whose centres are on the x-axis.

  1. Define the general equation of the circle.
  2. Find the DE corresponding to the above equation. 

1 Answer

+1 vote
by (28.9k points)
selected by
 
Best answer

 1. The general equation of the circle, passing through the origin and whose centers lies on x-axis can be taken as (x – h)2 + y2 = h2 where h being an arbitrary constant.

2. Simplifying (x – h)2 + y2 = h2 we get,

x2 – 2hx + h2 + y2 = h2 ⇒ x2 – 2hx + h2 = 0 _____(1)

Differentiating we get,

2x + 2y \(\frac{dy}{dx}\)  – 2h = 0 ⇒ h = x + y \(\frac{dy}{dx}\)

Substituting in (1) we can eliminate h


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.

...