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Circle with centre (0, 4) and passing through the projection of (2, 4) on x-axis is
- a)x2 + y2 -8y - 4 = 0
- b)x2 + y2 - 4x -8y + 4 = 0
- c)x2 + y2 - 8y + 4 = 0
- d)x2 + y2 - 8y +16 = 0
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Circle with centre (0, 4) and passing through the projection of (2, 4)...
Centre (0, 4) and r = 1
Most Upvoted Answer
Circle with centre (0, 4) and passing through the projection of (2, 4)...
Solution:
Given, the centre of the circle is (0, 4) and it passes through the projection of (2, 4) on x-axis.
Let the projection of (2, 4) on x-axis be (2, 0).
We know that the distance between the centre of the circle and any point on the circle is the radius of the circle.
Radius of the circle = Distance between (0, 4) and (2, 0)
= √((2-0)² + (0-4)²)
= √20
= 2√5
So, the equation of the circle can be written as:
(x-0)² + (y-4)² = (2√5)²
x² + y² - 8y + 16 - 20 = 0
x² + y² - 8y - 4 = 0
Hence, the correct option is (a) x² + y² - 8y - 4 = 0.
Given, the centre of the circle is (0, 4) and it passes through the projection of (2, 4) on x-axis.
Let the projection of (2, 4) on x-axis be (2, 0).
We know that the distance between the centre of the circle and any point on the circle is the radius of the circle.
Radius of the circle = Distance between (0, 4) and (2, 0)
= √((2-0)² + (0-4)²)
= √20
= 2√5
So, the equation of the circle can be written as:
(x-0)² + (y-4)² = (2√5)²
x² + y² - 8y + 16 - 20 = 0
x² + y² - 8y - 4 = 0
Hence, the correct option is (a) x² + y² - 8y - 4 = 0.
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Circle with centre (0, 4) and passing through the projection of (2, 4) on x-axis isa)x2 + y2 -8y - 4 = 0b)x2 + y2 - 4x -8y + 4 = 0c)x2 + y2 - 8y + 4 = 0d)x2 + y2 - 8y +16 = 0Correct answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Circle with centre (0, 4) and passing through the projection of (2, 4) on x-axis isa)x2 + y2 -8y - 4 = 0b)x2 + y2 - 4x -8y + 4 = 0c)x2 + y2 - 8y + 4 = 0d)x2 + y2 - 8y +16 = 0Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Circle with centre (0, 4) and passing through the projection of (2, 4) on x-axis isa)x2 + y2 -8y - 4 = 0b)x2 + y2 - 4x -8y + 4 = 0c)x2 + y2 - 8y + 4 = 0d)x2 + y2 - 8y +16 = 0Correct answer is option 'A'. Can you explain this answer?.
Circle with centre (0, 4) and passing through the projection of (2, 4) on x-axis isa)x2 + y2 -8y - 4 = 0b)x2 + y2 - 4x -8y + 4 = 0c)x2 + y2 - 8y + 4 = 0d)x2 + y2 - 8y +16 = 0Correct answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Circle with centre (0, 4) and passing through the projection of (2, 4) on x-axis isa)x2 + y2 -8y - 4 = 0b)x2 + y2 - 4x -8y + 4 = 0c)x2 + y2 - 8y + 4 = 0d)x2 + y2 - 8y +16 = 0Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Circle with centre (0, 4) and passing through the projection of (2, 4) on x-axis isa)x2 + y2 -8y - 4 = 0b)x2 + y2 - 4x -8y + 4 = 0c)x2 + y2 - 8y + 4 = 0d)x2 + y2 - 8y +16 = 0Correct answer is option 'A'. Can you explain this answer?.
Solutions for Circle with centre (0, 4) and passing through the projection of (2, 4) on x-axis isa)x2 + y2 -8y - 4 = 0b)x2 + y2 - 4x -8y + 4 = 0c)x2 + y2 - 8y + 4 = 0d)x2 + y2 - 8y +16 = 0Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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Circle with centre (0, 4) and passing through the projection of (2, 4) on x-axis isa)x2 + y2 -8y - 4 = 0b)x2 + y2 - 4x -8y + 4 = 0c)x2 + y2 - 8y + 4 = 0d)x2 + y2 - 8y +16 = 0Correct answer is option 'A'. Can you explain this answer?, a detailed solution for Circle with centre (0, 4) and passing through the projection of (2, 4) on x-axis isa)x2 + y2 -8y - 4 = 0b)x2 + y2 - 4x -8y + 4 = 0c)x2 + y2 - 8y + 4 = 0d)x2 + y2 - 8y +16 = 0Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of Circle with centre (0, 4) and passing through the projection of (2, 4) on x-axis isa)x2 + y2 -8y - 4 = 0b)x2 + y2 - 4x -8y + 4 = 0c)x2 + y2 - 8y + 4 = 0d)x2 + y2 - 8y +16 = 0Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
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