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JEE Mains Exam Centres in Telangana 2026- NTA released the JEE Main 2026 exam centres for Telangana on the official website, jeemain.nta.ac.in. Only candidates whose permanent state of residence is Telangana can opt for Telangana exam centres. The allotted exam centre is mentioned in the JEE Main admit card, which has been released. There were a total of 14 exam centres in cities across Telangana for the previous exam. Candidates can find the names and addresses of the test centres online. The authority will conduct JEE Main 2026 for admission to various BTech. and B.Arch courses at several prestigious colleges.
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.
The JEE Main city slip 2026 was released on January 8, 2026. Applicants must go through this article for the detailed list of Telangana exam centres for JEE Main 2026. The authority will conduct the JEE Main 2026 session 1 exam between January 21 to 29, 2026.
Also Check: JEE Main 2026 Exam Centres List
Students planning to take the JEE Main 2026 exam in Telangana will find the district, city, and city codes on their official website. As per this year's information brochure, JEE Main test centres are available in 11 district headquarters. Key cities in Telangana hosting JEE Main exam centres include Hyderabad, Karimnagar, Kothagudem, Nizamabad, Mahbubnagar, and others. Until then, candidates can check the previous year's JEE Main exam centres in Telangana.
Telangana | Karimnagar | TL02 |
Telangana | Khammam | TL03 |
Telangana | Mahbubnagar | TL04 |
Telangana | Nalgonda | TL05 |
Telangana | Warangal | TL07 |
Telangana | Nizamabad | TL08 |
Telangana | Suryapet | TL09 |
Telangana | Siddipet | TL11 |
Telangana | Jagtial | TL15 |
Telangana | Kothagudem | TL17 |
Telangana | Hyderabad/Secunderabad | TL01 |
Telangana | Adilabad | TL12 |
Telangana | Kodad | TL25 |
Telangana | Peddapalli | TL26 |
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Candidates with queries like “how to check my JEE Main exam centre” must note that the JEE Main test centre details are available on the admit card only. JEE Main hall ticket mentions the date, JEE Mains exam centres in Telangana 2026 with timings and much more.
Frequently Asked Questions (FAQs)
JEE Main 2026 exam dates are out. JEE Main 2026 session 1 exam will be held between January 21 to 29, and Session 2 will be held between April 2 to 9, 2026.
Candidates can check the JEE Main exam centres on the official website, jeemain.nta.nic.in. To know the assigned exam centre details check the admit card of JEE Main 2026. However, JEE city slip will provide information on the exam city and exam date.
On Question asked by student community
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-
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 Lucky,
Please refer to this link and you can download the free pdf.
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
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