JEE Exam > JEE Questions > The equation of circle whose centre is (2, 1)...
Start Learning for Free
The equation of circle whose centre is (2, 1) and which passes through the point (3, – 5) is:
- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
The equation of circle whose centre is (2, 1) and which passes through...
Radius of circle is given by -
r = √[(h-x1)² + (k-y1)²]
r = √[(2-3)² + (1+5)²]
r = √(-1² + 6²)
r = √(1 + 36)
r = √37
if centre (2,-]1) and radius=√26 are given,
(x-h)2+(y-k)2=r2
equation is (x-2)2 + (y-1)2 = (√37)2
x2 + 4 - 4x + y2 + 1 - 2y = 37
x2 + y2 - 4x - 2y - 32 = 0
r = √[(h-x1)² + (k-y1)²]
r = √[(2-3)² + (1+5)²]
r = √(-1² + 6²)
r = √(1 + 36)
r = √37
if centre (2,-]1) and radius=√26 are given,
(x-h)2+(y-k)2=r2
equation is (x-2)2 + (y-1)2 = (√37)2
x2 + 4 - 4x + y2 + 1 - 2y = 37
x2 + y2 - 4x - 2y - 32 = 0
Most Upvoted Answer
The equation of circle whose centre is (2, 1) and which passes through...
Radius of circle is given by -
r = √[(h-x1)² + (k-y1)²]
r = √[(2-3)² + (1+5)²]
r = √(-1² + 6²)
r = √(1 + 36)
r = √37
if centre (2,-]1) and radius=√26 are given,
(x-h)2+(y-k)2=r2
equation is (x-2)2 + (y-1)2 = (√37)2
x2 + 4 - 4x + y2 + 1 - 2y = 37
x2 + y2 - 4x - 2y - 32 = 0
r = √[(h-x1)² + (k-y1)²]
r = √[(2-3)² + (1+5)²]
r = √(-1² + 6²)
r = √(1 + 36)
r = √37
if centre (2,-]1) and radius=√26 are given,
(x-h)2+(y-k)2=r2
equation is (x-2)2 + (y-1)2 = (√37)2
x2 + 4 - 4x + y2 + 1 - 2y = 37
x2 + y2 - 4x - 2y - 32 = 0
Community Answer
The equation of circle whose centre is (2, 1) and which passes through...
The equation of a circle with center (h, k) and radius r is given by:
(x - h)^2 + (y - k)^2 = r^2
In this case, the center is (2, 1) and the circle passes through the point (3, 7). We can substitute these values into the equation to solve for the radius:
(3 - 2)^2 + (7 - 1)^2 = r^2
1 + 36 = r^2
37 = r^2
So, the equation of the circle is:
(x - 2)^2 + (y - 1)^2 = 37
(x - h)^2 + (y - k)^2 = r^2
In this case, the center is (2, 1) and the circle passes through the point (3, 7). We can substitute these values into the equation to solve for the radius:
(3 - 2)^2 + (7 - 1)^2 = r^2
1 + 36 = r^2
37 = r^2
So, the equation of the circle is:
(x - 2)^2 + (y - 1)^2 = 37
Attention JEE Students!
To make sure you are not studying endlessly, EduRev has designed JEE study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in JEE.
|
Explore Courses for JEE exam
|
|
Similar JEE Doubts
Top Courses for JEEView all
The equation of circle whose centre is (2, 1) and which passes through the point (3, – 5) is:a)b)c)d)Correct answer is option 'B'. Can you explain this answer?
Question Description
The equation of circle whose centre is (2, 1) and which passes through the point (3, – 5) is:a)b)c)d)Correct 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 The equation of circle whose centre is (2, 1) and which passes through the point (3, – 5) is:a)b)c)d)Correct 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 The equation of circle whose centre is (2, 1) and which passes through the point (3, – 5) is:a)b)c)d)Correct answer is option 'B'. Can you explain this answer?.
The equation of circle whose centre is (2, 1) and which passes through the point (3, – 5) is:a)b)c)d)Correct 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 The equation of circle whose centre is (2, 1) and which passes through the point (3, – 5) is:a)b)c)d)Correct 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 The equation of circle whose centre is (2, 1) and which passes through the point (3, – 5) is:a)b)c)d)Correct answer is option 'B'. Can you explain this answer?.
Solutions for The equation of circle whose centre is (2, 1) and which passes through the point (3, – 5) is:a)b)c)d)Correct 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 The equation of circle whose centre is (2, 1) and which passes through the point (3, – 5) is:a)b)c)d)Correct answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
The equation of circle whose centre is (2, 1) and which passes through the point (3, – 5) is:a)b)c)d)Correct answer is option 'B'. Can you explain this answer?, a detailed solution for The equation of circle whose centre is (2, 1) and which passes through the point (3, – 5) is:a)b)c)d)Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of The equation of circle whose centre is (2, 1) and which passes through the point (3, – 5) is:a)b)c)d)Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice The equation of circle whose centre is (2, 1) and which passes through the point (3, – 5) is:a)b)c)d)Correct answer is option 'B'. Can you explain this answer? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up
within 7 days!
within 7 days!
Takes less than 10 seconds to signup