JEE Exam > JEE Questions > The equation of the circle having its centre ...
Start Learning for Free
The equation of the circle having its centre on the line x + 2y - 3 = 0 and passing through the points of intersection of the circles x2+ y2 - 2x - 4y + 1 = 0 and x2 + y2 - 4x - 2y + 4 = 0 is
- a)x2 + y2 + 2x - 4y + 4 = 0
- b)x2 + y2 - 2x - 2y + 1 = 0
- c)x2 + y2 - 6x + 7 = 0
- d)x2 + y2 - 3x + 4 = 0
Correct answer is option 'C'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Most Upvoted Answer
The equation of the circle having its centre on the line x + 2y - 3 = ...
Solution:
Finding the Center of the Circle:
We are given that the center of the circle lies on the line x - 2y - 3 = 0. Let's find the coordinates of the center.
- The given line can be written in the standard form as, x - 2y = 3.
- Let's assume that the center of the circle is (h, k). Then the equation of the line passing through (h, k) and perpendicular to the given line is given by:
(y - k) = (1/2)(x - h)
- Simplifying this equation, we get: x - 2y - (h - 2k) = 0
- Since this line passes through (h, k), we have: h - 2k = 3
- We have two equations in two variables: x - 2y = 3 and h - 2k = 3. Solving these equations, we get: h = 3, k = 0.
- Therefore, the center of the circle is (3, 0).
Finding the Radius of the Circle:
We are given that the circle passes through the points of intersection of the circles x^2 + y^2 - 2x - 4y + 1 = 0 and x^2 + y^2 - 4x - 2y + 4 = 0. Let's find these points of intersection.
- Subtracting the second equation from the first, we get: 2x + 2y - 3 = 0.
- Simplifying this equation, we get: y = -(x/2) + 3/2.
- Substituting this value of y in the first equation and simplifying, we get: x^2 - 2x + 1 = 0.
- Solving this quadratic equation, we get: x = 1 (repeated root).
- Substituting this value of x in y = -(x/2) + 3/2, we get: y = 1.
- Therefore, the two points of intersection are (1, 1) and (1, -1).
- The distance between the center of the circle (3, 0) and either of these points of intersection is the radius of the circle. Using the distance formula, we get: r = 2.
Writing the Equation of the Circle:
- The equation of a circle with center (h, k) and radius r is given by: (x - h)^2 + (y - k)^2 = r^2.
- Substituting the values of (h, k, r), we get: (x - 3)^2 + y^2 = 4.
- Simplifying this equation, we get: x^2 + y^2 - 6x + 9 = 0.
- Subtracting 9 from both sides, we get: x^2 + y^2 - 6x = -9.
- Adding 9 and 36 to both sides (to complete the square), we get: x^2 - 6x + 9 + y^2 = 27.
- Simplifying this equation, we get: (x - 3)^2 + y^2 = 27.
- Therefore, the equation of the circle is: x^
Finding the Center of the Circle:
We are given that the center of the circle lies on the line x - 2y - 3 = 0. Let's find the coordinates of the center.
- The given line can be written in the standard form as, x - 2y = 3.
- Let's assume that the center of the circle is (h, k). Then the equation of the line passing through (h, k) and perpendicular to the given line is given by:
(y - k) = (1/2)(x - h)
- Simplifying this equation, we get: x - 2y - (h - 2k) = 0
- Since this line passes through (h, k), we have: h - 2k = 3
- We have two equations in two variables: x - 2y = 3 and h - 2k = 3. Solving these equations, we get: h = 3, k = 0.
- Therefore, the center of the circle is (3, 0).
Finding the Radius of the Circle:
We are given that the circle passes through the points of intersection of the circles x^2 + y^2 - 2x - 4y + 1 = 0 and x^2 + y^2 - 4x - 2y + 4 = 0. Let's find these points of intersection.
- Subtracting the second equation from the first, we get: 2x + 2y - 3 = 0.
- Simplifying this equation, we get: y = -(x/2) + 3/2.
- Substituting this value of y in the first equation and simplifying, we get: x^2 - 2x + 1 = 0.
- Solving this quadratic equation, we get: x = 1 (repeated root).
- Substituting this value of x in y = -(x/2) + 3/2, we get: y = 1.
- Therefore, the two points of intersection are (1, 1) and (1, -1).
- The distance between the center of the circle (3, 0) and either of these points of intersection is the radius of the circle. Using the distance formula, we get: r = 2.
Writing the Equation of the Circle:
- The equation of a circle with center (h, k) and radius r is given by: (x - h)^2 + (y - k)^2 = r^2.
- Substituting the values of (h, k, r), we get: (x - 3)^2 + y^2 = 4.
- Simplifying this equation, we get: x^2 + y^2 - 6x + 9 = 0.
- Subtracting 9 from both sides, we get: x^2 + y^2 - 6x = -9.
- Adding 9 and 36 to both sides (to complete the square), we get: x^2 - 6x + 9 + y^2 = 27.
- Simplifying this equation, we get: (x - 3)^2 + y^2 = 27.
- Therefore, the equation of the circle is: x^
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 the circle having its centre on the line x + 2y - 3 = 0 and passing through the points of intersection of the circles x2+ y2 - 2x - 4y + 1 = 0 and x2 + y2 - 4x - 2y + 4 = 0 isa)x2 + y2 + 2x - 4y + 4 = 0b)x2 + y2 - 2x - 2y + 1 = 0c)x2 + y2 - 6x + 7 = 0d)x2 + y2 - 3x + 4 = 0Correct answer is option 'C'. Can you explain this answer?
Question Description
The equation of the circle having its centre on the line x + 2y - 3 = 0 and passing through the points of intersection of the circles x2+ y2 - 2x - 4y + 1 = 0 and x2 + y2 - 4x - 2y + 4 = 0 isa)x2 + y2 + 2x - 4y + 4 = 0b)x2 + y2 - 2x - 2y + 1 = 0c)x2 + y2 - 6x + 7 = 0d)x2 + y2 - 3x + 4 = 0Correct 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 equation of the circle having its centre on the line x + 2y - 3 = 0 and passing through the points of intersection of the circles x2+ y2 - 2x - 4y + 1 = 0 and x2 + y2 - 4x - 2y + 4 = 0 isa)x2 + y2 + 2x - 4y + 4 = 0b)x2 + y2 - 2x - 2y + 1 = 0c)x2 + y2 - 6x + 7 = 0d)x2 + y2 - 3x + 4 = 0Correct 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 equation of the circle having its centre on the line x + 2y - 3 = 0 and passing through the points of intersection of the circles x2+ y2 - 2x - 4y + 1 = 0 and x2 + y2 - 4x - 2y + 4 = 0 isa)x2 + y2 + 2x - 4y + 4 = 0b)x2 + y2 - 2x - 2y + 1 = 0c)x2 + y2 - 6x + 7 = 0d)x2 + y2 - 3x + 4 = 0Correct answer is option 'C'. Can you explain this answer?.
The equation of the circle having its centre on the line x + 2y - 3 = 0 and passing through the points of intersection of the circles x2+ y2 - 2x - 4y + 1 = 0 and x2 + y2 - 4x - 2y + 4 = 0 isa)x2 + y2 + 2x - 4y + 4 = 0b)x2 + y2 - 2x - 2y + 1 = 0c)x2 + y2 - 6x + 7 = 0d)x2 + y2 - 3x + 4 = 0Correct 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 equation of the circle having its centre on the line x + 2y - 3 = 0 and passing through the points of intersection of the circles x2+ y2 - 2x - 4y + 1 = 0 and x2 + y2 - 4x - 2y + 4 = 0 isa)x2 + y2 + 2x - 4y + 4 = 0b)x2 + y2 - 2x - 2y + 1 = 0c)x2 + y2 - 6x + 7 = 0d)x2 + y2 - 3x + 4 = 0Correct 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 equation of the circle having its centre on the line x + 2y - 3 = 0 and passing through the points of intersection of the circles x2+ y2 - 2x - 4y + 1 = 0 and x2 + y2 - 4x - 2y + 4 = 0 isa)x2 + y2 + 2x - 4y + 4 = 0b)x2 + y2 - 2x - 2y + 1 = 0c)x2 + y2 - 6x + 7 = 0d)x2 + y2 - 3x + 4 = 0Correct answer is option 'C'. Can you explain this answer?.
Solutions for The equation of the circle having its centre on the line x + 2y - 3 = 0 and passing through the points of intersection of the circles x2+ y2 - 2x - 4y + 1 = 0 and x2 + y2 - 4x - 2y + 4 = 0 isa)x2 + y2 + 2x - 4y + 4 = 0b)x2 + y2 - 2x - 2y + 1 = 0c)x2 + y2 - 6x + 7 = 0d)x2 + y2 - 3x + 4 = 0Correct 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 equation of the circle having its centre on the line x + 2y - 3 = 0 and passing through the points of intersection of the circles x2+ y2 - 2x - 4y + 1 = 0 and x2 + y2 - 4x - 2y + 4 = 0 isa)x2 + y2 + 2x - 4y + 4 = 0b)x2 + y2 - 2x - 2y + 1 = 0c)x2 + y2 - 6x + 7 = 0d)x2 + y2 - 3x + 4 = 0Correct answer is option 'C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
The equation of the circle having its centre on the line x + 2y - 3 = 0 and passing through the points of intersection of the circles x2+ y2 - 2x - 4y + 1 = 0 and x2 + y2 - 4x - 2y + 4 = 0 isa)x2 + y2 + 2x - 4y + 4 = 0b)x2 + y2 - 2x - 2y + 1 = 0c)x2 + y2 - 6x + 7 = 0d)x2 + y2 - 3x + 4 = 0Correct answer is option 'C'. Can you explain this answer?, a detailed solution for The equation of the circle having its centre on the line x + 2y - 3 = 0 and passing through the points of intersection of the circles x2+ y2 - 2x - 4y + 1 = 0 and x2 + y2 - 4x - 2y + 4 = 0 isa)x2 + y2 + 2x - 4y + 4 = 0b)x2 + y2 - 2x - 2y + 1 = 0c)x2 + y2 - 6x + 7 = 0d)x2 + y2 - 3x + 4 = 0Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of The equation of the circle having its centre on the line x + 2y - 3 = 0 and passing through the points of intersection of the circles x2+ y2 - 2x - 4y + 1 = 0 and x2 + y2 - 4x - 2y + 4 = 0 isa)x2 + y2 + 2x - 4y + 4 = 0b)x2 + y2 - 2x - 2y + 1 = 0c)x2 + y2 - 6x + 7 = 0d)x2 + y2 - 3x + 4 = 0Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice The equation of the circle having its centre on the line x + 2y - 3 = 0 and passing through the points of intersection of the circles x2+ y2 - 2x - 4y + 1 = 0 and x2 + y2 - 4x - 2y + 4 = 0 isa)x2 + y2 + 2x - 4y + 4 = 0b)x2 + y2 - 2x - 2y + 1 = 0c)x2 + y2 - 6x + 7 = 0d)x2 + y2 - 3x + 4 = 0Correct 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