Find the equation of the circle passing through the points (2, 3) and (–1, 1) and whose centre is on the line x – 3y – 11 = 0.

0 votes
12 views
asked Feb 9, 2018 in Mathematics by Rohit Singh (61,782 points) 35 133 357

Find the equation of the circle passing through the points (2, 3) and (–1, 1) and whose centre is on the line x – 3y – 11 = 0.

1 Answer

0 votes
answered Feb 9, 2018 by sameer (82,980 points) 5 11 37

Let the equation of the required circle be (x – h)2 + (y – k)2 = r2.
Since the circle passes through points (2, 3) and (–1, 1),
(2 – h)2 + (3 – k)2 = r2 ………………………….. (1)
(–1 – h)2 + (1 – k)2 = r2 ………………………… (2)
Since the centre (h, k) of the circle lies on line x – 3y – 11 = 0,
h – 3k = 11 …………………………………………….. (3)
From equations (1) and (2), we obtain
(2 – h)2 + (3 – k)2 = (–1 – h)2 + (1 – k)2
⇒ 4 – 4h + h2 + 9 – 6k + k2 = 1 + 2h + h2 + 1 – 2k + k2
⇒ 4 – 4h + 9 – 6k = 1 + 2h + 1 – 2k
⇒ 6h + 4k = 11 … (4)
On solving equations (3) and (4), we obtain ℎ=7/2 and =−5/2.
On substituting the values of h and k in equation (1), we obtain

Thus, the equation of the required circle is

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.

One Thought Forever

“There is a close connection between getting up in the world and getting up in the morning.“
– Anon
~~~*****~~~

...