JEE Main 2026 Exam Centres in Chennai

JEE Main 2026 Exam Centres in Chennai

Team Careers360Updated on 18 Jan 2026, 05:30 PM IST

JEE Main 2026 Exam Centres in Chennai - NTA has released the JEE Mains 2026 exam centres in Chennai on the official website, jeemain.nta.nic.in. Along with the brochure, the NTA has released the JEE Main exam centres 2026. Candidates can check the JEE Main Chennai exam centres list in this article. The JEE Main exam centres will help aspirants navigate the designated physical locations where the exam will be held. The exam centres are released a few weeks before the exam. The JEE Main exam is held online mode at test centres. Candidates can check the JEE Main admit card to know the exact address of the JEE Main test centre in Chennai. National Testing Agency is conducting the JEE Main 2026 exam for two sessions in January & April 2026.

LiveJEE Main 2026 LIVE: NTA to begin session 2 registration soon; shift-wise exam analysisJan 28, 2026 | 10:07 PM IST

For JEE Main 2026 session 2, candidates who passed the Class 12 or equivalent examination in 2023 or earlier are not eligible to appear for JEE Main 2026 session 2. The passing year is counted as the year in which a candidate is declared ‘pass’ for the first time in Class 12. Subsequent attempts, subject additions, or improvement exams will not be considered as a fresh passing year for eligibility purposes.

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JEE Main 2026 Exam Centres in Chennai
JEE Main 2026 Exam Centres in Chennai

JEE Main 2026 Exam Centres in Chennai With Address

NTA will release the JEE Main exam centres in Chennai 2026. Students who want to take the exam in Chennai can check the list of available IIT JEE Test centres here. The test city code for Chennai is TN01. Find below the JEE Main exam centres in Chennai in the table below.

JEE Main Exam Centres in Chennai

Name of Exam Centre

Address of Exam Centre

iON Digital Zone

Dhanalakshmi College of Engineering Manimangalam Post, Tambaram, Chennai, Tamil Nadu 601301

TCS iON Digital Zone

1st, 2nd Floors, Fortune Towers SH 109, 2nd St, Ganapathy Nagar, Kovilambakkam, Chennai, Tamil Nadu 600117

Ion Digital Zone

Kundrathur, Mangadu Rd, Sarathy Nagar, Chennai, Tamil Nadu 600069

JEE Main Exam Centres List

Frequently Asked Questions (FAQs)

Q: How do I check the JEE Main exam centres in Chennai?
A:

Candidates can check the location of their exam centre in Chennai from the official website jeemain.nta.nic.in.

Q: Can I apply for changing JEE Main exam centre?
A:

Yes, after filling out the application form but it is not possible after the release of the admit card.

Q: What document is required to be carried at JEE Main exam centres in Chennai?
A:

Candidates need to carry their admit card along with a passport-size photograph and a valid ID proof.

Q: When will the authorities release the JEE Main admit card?
A:

The authority has released the JEE Main admit card for session 1 on January 17, 2026.

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Questions related to JEE Main

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Have a question related to JEE Main ?

JEE Main 28 Jan shift 2 exam will end at 6 PM. The complete analysis and memory-based questions will solution will be updated in the below article. Keep checking the page-

https://engineering.careers360.com/articles/jee-main-2026-january-28-shift-2-question-paper-with-solutions-pdf

Math and Chemistry was difficult and Physics was moderate. The complete analysis is available here- https://engineering.careers360.com/articles/jee-main-2026-january-28-shift-1-question-paper-with-solutions-pdf

You can also check the memory-based questions and detailed solutions for JEE Main Jan 28 shift 1 paper.

HI Manisha Maharana

You can download the JEE Mains 10 Free Mock Test with Detailed solutions. Its a feely downloadable pdf.
https://engineering.careers360.com/download/sample-papers/jee-main-10-full-mock-test-and-explanations-pdf

Also, you can check ad attemp the online mock test on our platform.
https://learn.careers360.com/test-series-jee-main-free-mock-test/



A general equation of a circle is

$
x^2+y^2+2 g x+2 f y+c=0
$

Since it passes through $(0,0)$,

$
c=0
$

So the equation becomes

$
x^2+y^2+2 g x+2 f y=0
$

It cuts the x -axis at ( $a, 0$ ).

Substituting:

$
a^2+2 g a=0
$

g=-a/2