JEE Exam > JEE Questions > The parametric coordinates of any point on th...
Start Learning for Free
The parametric coordinates of any point on the circle x2 + y2 – 4x – 4y = 0 are
- a)(–2 + 2cosa, –2 + 2 sin a)
- b)(2 + 2cosa, 4 + 2 sina)
- c)(2 + 2√2cosa, 2 + 2√2sina)
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
The parametric coordinates of any point on the circle x2+ y2–4x ...
Most Upvoted Answer
The parametric coordinates of any point on the circle x2+ y2–4x ...
To find the parametric coordinates of any point on the circle x^2 + y^2 – 4x – 4y = 0, we can use the concept of the unit circle and trigonometry.
Given equation: x^2 + y^2 – 4x – 4y = 0
Completing the square for x and y, we get:
(x^2 – 4x) + (y^2 – 4y) = 0
(x^2 – 4x + 4) + (y^2 – 4y + 4) = 8
(x – 2)^2 + (y – 2)^2 = 8
Comparing this equation with the standard equation of a circle, (x – h)^2 + (y – k)^2 = r^2, we can see that the center of the circle is at (2, 2) and the radius is √8 = 2√2.
Since the equation is in standard form, we can parameterize the circle using trigonometric functions.
Let x = 2 + 2√2cos(a) and y = 2 + 2√2sin(a), where a is the parameter.
Substituting these values into the equation, we get:
(2 + 2√2cos(a) – 2)^2 + (2 + 2√2sin(a) – 2)^2 = 8
(2√2cos(a))^2 + (2√2sin(a))^2 = 8
8cos^2(a) + 8sin^2(a) = 8
8(cos^2(a) + sin^2(a)) = 8
8 = 8
Since the equation is satisfied, the parameterized coordinates (2 + 2√2cos(a), 2 + 2√2sin(a)) lie on the circle x^2 + y^2 – 4x – 4y = 0.
Therefore, the correct answer is option C: (2 + 2√2cos(a), 2 + 2√2sin(a)).
Given equation: x^2 + y^2 – 4x – 4y = 0
Completing the square for x and y, we get:
(x^2 – 4x) + (y^2 – 4y) = 0
(x^2 – 4x + 4) + (y^2 – 4y + 4) = 8
(x – 2)^2 + (y – 2)^2 = 8
Comparing this equation with the standard equation of a circle, (x – h)^2 + (y – k)^2 = r^2, we can see that the center of the circle is at (2, 2) and the radius is √8 = 2√2.
Since the equation is in standard form, we can parameterize the circle using trigonometric functions.
Let x = 2 + 2√2cos(a) and y = 2 + 2√2sin(a), where a is the parameter.
Substituting these values into the equation, we get:
(2 + 2√2cos(a) – 2)^2 + (2 + 2√2sin(a) – 2)^2 = 8
(2√2cos(a))^2 + (2√2sin(a))^2 = 8
8cos^2(a) + 8sin^2(a) = 8
8(cos^2(a) + sin^2(a)) = 8
8 = 8
Since the equation is satisfied, the parameterized coordinates (2 + 2√2cos(a), 2 + 2√2sin(a)) lie on the circle x^2 + y^2 – 4x – 4y = 0.
Therefore, the correct answer is option C: (2 + 2√2cos(a), 2 + 2√2sin(a)).
Community Answer
The parametric coordinates of any point on the circle x2+ y2–4x ...
Solution to this que can be obtained by solving the equation by the standard equation and finding the centre co ordinates and the r radius of the circle.By this we get the parametric equations.And the given option C is the correct one.
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 parametric coordinates of any point on the circle x2+ y2–4x –4y = 0 area)(–2 + 2cosa, –2 + 2 sin a)b)(2 + 2cosa, 4 + 2 sina)c)(2 + 2√2cosa, 2 + 2√2sina)d)None of theseCorrect answer is option 'C'. Can you explain this answer?
Question Description
The parametric coordinates of any point on the circle x2+ y2–4x –4y = 0 area)(–2 + 2cosa, –2 + 2 sin a)b)(2 + 2cosa, 4 + 2 sina)c)(2 + 2√2cosa, 2 + 2√2sina)d)None of theseCorrect answer is option 'C'. 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 parametric coordinates of any point on the circle x2+ y2–4x –4y = 0 area)(–2 + 2cosa, –2 + 2 sin a)b)(2 + 2cosa, 4 + 2 sina)c)(2 + 2√2cosa, 2 + 2√2sina)d)None of theseCorrect answer is option 'C'. 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 parametric coordinates of any point on the circle x2+ y2–4x –4y = 0 area)(–2 + 2cosa, –2 + 2 sin a)b)(2 + 2cosa, 4 + 2 sina)c)(2 + 2√2cosa, 2 + 2√2sina)d)None of theseCorrect answer is option 'C'. Can you explain this answer?.
The parametric coordinates of any point on the circle x2+ y2–4x –4y = 0 area)(–2 + 2cosa, –2 + 2 sin a)b)(2 + 2cosa, 4 + 2 sina)c)(2 + 2√2cosa, 2 + 2√2sina)d)None of theseCorrect answer is option 'C'. 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 parametric coordinates of any point on the circle x2+ y2–4x –4y = 0 area)(–2 + 2cosa, –2 + 2 sin a)b)(2 + 2cosa, 4 + 2 sina)c)(2 + 2√2cosa, 2 + 2√2sina)d)None of theseCorrect answer is option 'C'. 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 parametric coordinates of any point on the circle x2+ y2–4x –4y = 0 area)(–2 + 2cosa, –2 + 2 sin a)b)(2 + 2cosa, 4 + 2 sina)c)(2 + 2√2cosa, 2 + 2√2sina)d)None of theseCorrect answer is option 'C'. Can you explain this answer?.
Solutions for The parametric coordinates of any point on the circle x2+ y2–4x –4y = 0 area)(–2 + 2cosa, –2 + 2 sin a)b)(2 + 2cosa, 4 + 2 sina)c)(2 + 2√2cosa, 2 + 2√2sina)d)None of theseCorrect answer is option 'C'. 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 parametric coordinates of any point on the circle x2+ y2–4x –4y = 0 area)(–2 + 2cosa, –2 + 2 sin a)b)(2 + 2cosa, 4 + 2 sina)c)(2 + 2√2cosa, 2 + 2√2sina)d)None of theseCorrect answer is option 'C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
The parametric coordinates of any point on the circle x2+ y2–4x –4y = 0 area)(–2 + 2cosa, –2 + 2 sin a)b)(2 + 2cosa, 4 + 2 sina)c)(2 + 2√2cosa, 2 + 2√2sina)d)None of theseCorrect answer is option 'C'. Can you explain this answer?, a detailed solution for The parametric coordinates of any point on the circle x2+ y2–4x –4y = 0 area)(–2 + 2cosa, –2 + 2 sin a)b)(2 + 2cosa, 4 + 2 sina)c)(2 + 2√2cosa, 2 + 2√2sina)d)None of theseCorrect answer is option 'C'. Can you explain this answer? has been provided alongside types of The parametric coordinates of any point on the circle x2+ y2–4x –4y = 0 area)(–2 + 2cosa, –2 + 2 sin a)b)(2 + 2cosa, 4 + 2 sina)c)(2 + 2√2cosa, 2 + 2√2sina)d)None of theseCorrect answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice The parametric coordinates of any point on the circle x2+ y2–4x –4y = 0 area)(–2 + 2cosa, –2 + 2 sin a)b)(2 + 2cosa, 4 + 2 sina)c)(2 + 2√2cosa, 2 + 2√2sina)d)None of theseCorrect answer is option 'C'. 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!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup