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Tangents drawn from the point P(1, 8) to the circle x2 + y2 –6x– 4y –11 = 0 touch the circle at the points A and B. The equation of the circumcircle of the triangle PAB is (2009)a)x2 + y2 + 4x– 6y +19 = 0b)x2 + y2 – 4x – 10y + 19 = 0c)x2 + y2 – 2x + 6y – 29 = 0d)x2 + y2 – 6x – 4y +19 = 0Correct answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Tangents drawn from the point P(1, 8) to the circle x2 + y2 –6x– 4y –11 = 0 touch the circle at the points A and B. The equation of the circumcircle of the triangle PAB is (2009)a)x2 + y2 + 4x– 6y +19 = 0b)x2 + y2 – 4x – 10y + 19 = 0c)x2 + y2 – 2x + 6y – 29 = 0d)x2 + y2 – 6x – 4y +19 = 0Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Tangents drawn from the point P(1, 8) to the circle x2 + y2 –6x– 4y –11 = 0 touch the circle at the points A and B. The equation of the circumcircle of the triangle PAB is (2009)a)x2 + y2 + 4x– 6y +19 = 0b)x2 + y2 – 4x – 10y + 19 = 0c)x2 + y2 – 2x + 6y – 29 = 0d)x2 + y2 – 6x – 4y +19 = 0Correct answer is option 'B'. Can you explain this answer?.

Solutions for Tangents drawn from the point P(1, 8) to the circle x2 + y2 –6x– 4y –11 = 0 touch the circle at the points A and B. The equation of the circumcircle of the triangle PAB is (2009)a)x2 + y2 + 4x– 6y +19 = 0b)x2 + y2 – 4x – 10y + 19 = 0c)x2 + y2 – 2x + 6y – 29 = 0d)x2 + y2 – 6x – 4y +19 = 0Correct answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.

Here you can find the meaning of Tangents drawn from the point P(1, 8) to the circle x2 + y2 –6x– 4y –11 = 0 touch the circle at the points A and B. The equation of the circumcircle of the triangle PAB is (2009)a)x2 + y2 + 4x– 6y +19 = 0b)x2 + y2 – 4x – 10y + 19 = 0c)x2 + y2 – 2x + 6y – 29 = 0d)x2 + y2 – 6x – 4y +19 = 0Correct answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of Tangents drawn from the point P(1, 8) to the circle x2 + y2 –6x– 4y –11 = 0 touch the circle at the points A and B. The equation of the circumcircle of the triangle PAB is (2009)a)x2 + y2 + 4x– 6y +19 = 0b)x2 + y2 – 4x – 10y + 19 = 0c)x2 + y2 – 2x + 6y – 29 = 0d)x2 + y2 – 6x – 4y +19 = 0Correct answer is option 'B'. Can you explain this answer?, a detailed solution for Tangents drawn from the point P(1, 8) to the circle x2 + y2 –6x– 4y –11 = 0 touch the circle at the points A and B. The equation of the circumcircle of the triangle PAB is (2009)a)x2 + y2 + 4x– 6y +19 = 0b)x2 + y2 – 4x – 10y + 19 = 0c)x2 + y2 – 2x + 6y – 29 = 0d)x2 + y2 – 6x – 4y +19 = 0Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of Tangents drawn from the point P(1, 8) to the circle x2 + y2 –6x– 4y –11 = 0 touch the circle at the points A and B. The equation of the circumcircle of the triangle PAB is (2009)a)x2 + y2 + 4x– 6y +19 = 0b)x2 + y2 – 4x – 10y + 19 = 0c)x2 + y2 – 2x + 6y – 29 = 0d)x2 + y2 – 6x – 4y +19 = 0Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice Tangents drawn from the point P(1, 8) to the circle x2 + y2 –6x– 4y –11 = 0 touch the circle at the points A and B. The equation of the circumcircle of the triangle PAB is (2009)a)x2 + y2 + 4x– 6y +19 = 0b)x2 + y2 – 4x – 10y + 19 = 0c)x2 + y2 – 2x + 6y – 29 = 0d)x2 + y2 – 6x – 4y +19 = 0Correct answer is option 'B'. Can you explain this answer? tests, examples and also practice JEE tests.