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If a circle passes through the point (a, b) and cuts the circle x2 + y2 = 4 orthogonally, then the locus of its centre is
-[AIEEE-2004]
- a)2ax + 2by + (a2 + b2 + 4) = 0
- b)2ax + 2by – (a2 + b2 + 4) = 0
- c)2ax – 2by + (a2 + b2 + 4) = 0
- d)2ax – 2by – (a2 + b2 + 4) = 0
Correct answer is option 'B'. Can you explain this answer?
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To solve this problem, we will use the concept of orthogonal circles.
First, let's consider the equation of the circle x^2 + y^2 = 4. This circle has its center at the origin (0, 0) and a radius of 2.
Now, let's assume that the given circle passes through the point (a, b). Since the given circle cuts the circle x^2 + y^2 = 4 orthogonally, the distance between the centers of the two circles must be equal to the sum of their radii.
Let the center of the given circle be (h, k). The distance between the centers is given by √((h-0)^2 + (k-0)^2) = √(h^2 + k^2).
The sum of the radii is equal to 2 + √(h^2 + k^2).
Therefore, we can write the equation √(h^2 + k^2) = 2 + √(h^2 + k^2).
Simplifying this equation, we get √(h^2 + k^2) - √(h^2 + k^2) = 2.
This equation simplifies to 0 = 2, which is not possible.
Therefore, there is no circle that passes through the point (a, b) and cuts the circle x^2 + y^2 = 4 orthogonally.
Hence, the locus of its center is a null set.
First, let's consider the equation of the circle x^2 + y^2 = 4. This circle has its center at the origin (0, 0) and a radius of 2.
Now, let's assume that the given circle passes through the point (a, b). Since the given circle cuts the circle x^2 + y^2 = 4 orthogonally, the distance between the centers of the two circles must be equal to the sum of their radii.
Let the center of the given circle be (h, k). The distance between the centers is given by √((h-0)^2 + (k-0)^2) = √(h^2 + k^2).
The sum of the radii is equal to 2 + √(h^2 + k^2).
Therefore, we can write the equation √(h^2 + k^2) = 2 + √(h^2 + k^2).
Simplifying this equation, we get √(h^2 + k^2) - √(h^2 + k^2) = 2.
This equation simplifies to 0 = 2, which is not possible.
Therefore, there is no circle that passes through the point (a, b) and cuts the circle x^2 + y^2 = 4 orthogonally.
Hence, the locus of its center is a null set.
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If a circle passes through the point (a, b) and cuts the circle x2 + y2 = 4 orthogonally, then the locus of its centre is-[AIEEE-2004]a)2ax + 2by + (a2 + b2 + 4) = 0b)2ax + 2by – (a2 + b2 + 4) = 0c)2ax – 2by + (a2 + b2 + 4) = 0d)2ax – 2by – (a2 + b2 + 4) = 0Correct answer is option 'B'. Can you explain this answer?
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If a circle passes through the point (a, b) and cuts the circle x2 + y2 = 4 orthogonally, then the locus of its centre is-[AIEEE-2004]a)2ax + 2by + (a2 + b2 + 4) = 0b)2ax + 2by – (a2 + b2 + 4) = 0c)2ax – 2by + (a2 + b2 + 4) = 0d)2ax – 2by – (a2 + b2 + 4) = 0Correct 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 If a circle passes through the point (a, b) and cuts the circle x2 + y2 = 4 orthogonally, then the locus of its centre is-[AIEEE-2004]a)2ax + 2by + (a2 + b2 + 4) = 0b)2ax + 2by – (a2 + b2 + 4) = 0c)2ax – 2by + (a2 + b2 + 4) = 0d)2ax – 2by – (a2 + b2 + 4) = 0Correct 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 If a circle passes through the point (a, b) and cuts the circle x2 + y2 = 4 orthogonally, then the locus of its centre is-[AIEEE-2004]a)2ax + 2by + (a2 + b2 + 4) = 0b)2ax + 2by – (a2 + b2 + 4) = 0c)2ax – 2by + (a2 + b2 + 4) = 0d)2ax – 2by – (a2 + b2 + 4) = 0Correct answer is option 'B'. Can you explain this answer?.
If a circle passes through the point (a, b) and cuts the circle x2 + y2 = 4 orthogonally, then the locus of its centre is-[AIEEE-2004]a)2ax + 2by + (a2 + b2 + 4) = 0b)2ax + 2by – (a2 + b2 + 4) = 0c)2ax – 2by + (a2 + b2 + 4) = 0d)2ax – 2by – (a2 + b2 + 4) = 0Correct 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 If a circle passes through the point (a, b) and cuts the circle x2 + y2 = 4 orthogonally, then the locus of its centre is-[AIEEE-2004]a)2ax + 2by + (a2 + b2 + 4) = 0b)2ax + 2by – (a2 + b2 + 4) = 0c)2ax – 2by + (a2 + b2 + 4) = 0d)2ax – 2by – (a2 + b2 + 4) = 0Correct 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 If a circle passes through the point (a, b) and cuts the circle x2 + y2 = 4 orthogonally, then the locus of its centre is-[AIEEE-2004]a)2ax + 2by + (a2 + b2 + 4) = 0b)2ax + 2by – (a2 + b2 + 4) = 0c)2ax – 2by + (a2 + b2 + 4) = 0d)2ax – 2by – (a2 + b2 + 4) = 0Correct answer is option 'B'. Can you explain this answer?.
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If a circle passes through the point (a, b) and cuts the circle x2 + y2 = 4 orthogonally, then the locus of its centre is-[AIEEE-2004]a)2ax + 2by + (a2 + b2 + 4) = 0b)2ax + 2by – (a2 + b2 + 4) = 0c)2ax – 2by + (a2 + b2 + 4) = 0d)2ax – 2by – (a2 + b2 + 4) = 0Correct answer is option 'B'. Can you explain this answer?, a detailed solution for If a circle passes through the point (a, b) and cuts the circle x2 + y2 = 4 orthogonally, then the locus of its centre is-[AIEEE-2004]a)2ax + 2by + (a2 + b2 + 4) = 0b)2ax + 2by – (a2 + b2 + 4) = 0c)2ax – 2by + (a2 + b2 + 4) = 0d)2ax – 2by – (a2 + b2 + 4) = 0Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of If a circle passes through the point (a, b) and cuts the circle x2 + y2 = 4 orthogonally, then the locus of its centre is-[AIEEE-2004]a)2ax + 2by + (a2 + b2 + 4) = 0b)2ax + 2by – (a2 + b2 + 4) = 0c)2ax – 2by + (a2 + b2 + 4) = 0d)2ax – 2by – (a2 + b2 + 4) = 0Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
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