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The equation of the circle described on the common chord of the circles x2 + y2 -8x + y -15 = 0 and x2 + y2 - 4x + 4y - 42 = 0 as diameter is
- a)x2 + y2- x + 2y + 4 = 0
- b)x2+ y2 + 10x - 2y + 12 = 0
- c)x2 + y2 - 12x - 2y + 12 = 0
- d)x2 + y2 - 5x + 3y + 7 = 0
Correct answer is option 'C'. Can you explain this answer?
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The equation of the circle described on the common chord of the circle...
Solution:
Given two circles are x^2 + y^2 - 8x - y -15 = 0 and x^2 + y^2 - 4x - 4y - 42 = 0.
Let the equation of the common chord of the given circles be ax + by + c = 0.
The equation of the circle with this line as diameter is given by
(x - h)^2 + (y - k)^2 = r^2, where
(h, k) = (-a/2, -b/2) is the midpoint of the chord,
r = (1/2) * √(a^2 + b^2) is the radius of the circle.
Now, let's find the equation of the common chord of the given circles.
Equation of the first circle: x^2 + y^2 - 8x - y - 15 = 0
Equation of the second circle: x^2 + y^2 - 4x - 4y - 42 = 0
Subtracting the second equation from the first equation, we get
-4x + 3y + 27 = 0
Therefore, the equation of the common chord is -4x + 3y + 27 = 0.
Now, we can find the equation of the circle with this line as diameter.
Midpoint of the chord: (h, k) = (2, -9/4)
Radius of the circle: r = (1/2) * √((-4)^2 + 3^2) = √13/2
Therefore, the equation of the circle is
(x - 2)^2 + (y + 9/4)^2 = 13/2
Simplifying the equation, we get
x^2 + y^2 - 12x - 2y + 12 = 0
Hence, the correct option is (C) x^2 + y^2 - 12x - 2y + 12 = 0.
Given two circles are x^2 + y^2 - 8x - y -15 = 0 and x^2 + y^2 - 4x - 4y - 42 = 0.
Let the equation of the common chord of the given circles be ax + by + c = 0.
The equation of the circle with this line as diameter is given by
(x - h)^2 + (y - k)^2 = r^2, where
(h, k) = (-a/2, -b/2) is the midpoint of the chord,
r = (1/2) * √(a^2 + b^2) is the radius of the circle.
Now, let's find the equation of the common chord of the given circles.
Equation of the first circle: x^2 + y^2 - 8x - y - 15 = 0
Equation of the second circle: x^2 + y^2 - 4x - 4y - 42 = 0
Subtracting the second equation from the first equation, we get
-4x + 3y + 27 = 0
Therefore, the equation of the common chord is -4x + 3y + 27 = 0.
Now, we can find the equation of the circle with this line as diameter.
Midpoint of the chord: (h, k) = (2, -9/4)
Radius of the circle: r = (1/2) * √((-4)^2 + 3^2) = √13/2
Therefore, the equation of the circle is
(x - 2)^2 + (y + 9/4)^2 = 13/2
Simplifying the equation, we get
x^2 + y^2 - 12x - 2y + 12 = 0
Hence, the correct option is (C) x^2 + y^2 - 12x - 2y + 12 = 0.
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The equation of the circle described on the common chord of the circles x2 + y2 -8x + y -15 = 0 and x2 + y2 - 4x + 4y - 42 = 0 as diameter isa)x2 + y2- x + 2y + 4 = 0b)x2+ y2 + 10x - 2y + 12 = 0c)x2 + y2 - 12x - 2y + 12 = 0d)x2+ y2 - 5x + 3y + 7 = 0Correct answer is option 'C'. 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 The equation of the circle described on the common chord of the circles x2 + y2 -8x + y -15 = 0 and x2 + y2 - 4x + 4y - 42 = 0 as diameter isa)x2 + y2- x + 2y + 4 = 0b)x2+ y2 + 10x - 2y + 12 = 0c)x2 + y2 - 12x - 2y + 12 = 0d)x2+ y2 - 5x + 3y + 7 = 0Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The equation of the circle described on the common chord of the circles x2 + y2 -8x + y -15 = 0 and x2 + y2 - 4x + 4y - 42 = 0 as diameter isa)x2 + y2- x + 2y + 4 = 0b)x2+ y2 + 10x - 2y + 12 = 0c)x2 + y2 - 12x - 2y + 12 = 0d)x2+ y2 - 5x + 3y + 7 = 0Correct answer is option 'C'. Can you explain this answer?.
The equation of the circle described on the common chord of the circles x2 + y2 -8x + y -15 = 0 and x2 + y2 - 4x + 4y - 42 = 0 as diameter isa)x2 + y2- x + 2y + 4 = 0b)x2+ y2 + 10x - 2y + 12 = 0c)x2 + y2 - 12x - 2y + 12 = 0d)x2+ y2 - 5x + 3y + 7 = 0Correct answer is option 'C'. 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 The equation of the circle described on the common chord of the circles x2 + y2 -8x + y -15 = 0 and x2 + y2 - 4x + 4y - 42 = 0 as diameter isa)x2 + y2- x + 2y + 4 = 0b)x2+ y2 + 10x - 2y + 12 = 0c)x2 + y2 - 12x - 2y + 12 = 0d)x2+ y2 - 5x + 3y + 7 = 0Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The equation of the circle described on the common chord of the circles x2 + y2 -8x + y -15 = 0 and x2 + y2 - 4x + 4y - 42 = 0 as diameter isa)x2 + y2- x + 2y + 4 = 0b)x2+ y2 + 10x - 2y + 12 = 0c)x2 + y2 - 12x - 2y + 12 = 0d)x2+ y2 - 5x + 3y + 7 = 0Correct answer is option 'C'. Can you explain this answer?.
Solutions for The equation of the circle described on the common chord of the circles x2 + y2 -8x + y -15 = 0 and x2 + y2 - 4x + 4y - 42 = 0 as diameter isa)x2 + y2- x + 2y + 4 = 0b)x2+ y2 + 10x - 2y + 12 = 0c)x2 + y2 - 12x - 2y + 12 = 0d)x2+ y2 - 5x + 3y + 7 = 0Correct answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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The equation of the circle described on the common chord of the circles x2 + y2 -8x + y -15 = 0 and x2 + y2 - 4x + 4y - 42 = 0 as diameter isa)x2 + y2- x + 2y + 4 = 0b)x2+ y2 + 10x - 2y + 12 = 0c)x2 + y2 - 12x - 2y + 12 = 0d)x2+ y2 - 5x + 3y + 7 = 0Correct answer is option 'C'. Can you explain this answer?, a detailed solution for The equation of the circle described on the common chord of the circles x2 + y2 -8x + y -15 = 0 and x2 + y2 - 4x + 4y - 42 = 0 as diameter isa)x2 + y2- x + 2y + 4 = 0b)x2+ y2 + 10x - 2y + 12 = 0c)x2 + y2 - 12x - 2y + 12 = 0d)x2+ y2 - 5x + 3y + 7 = 0Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of The equation of the circle described on the common chord of the circles x2 + y2 -8x + y -15 = 0 and x2 + y2 - 4x + 4y - 42 = 0 as diameter isa)x2 + y2- x + 2y + 4 = 0b)x2+ y2 + 10x - 2y + 12 = 0c)x2 + y2 - 12x - 2y + 12 = 0d)x2+ y2 - 5x + 3y + 7 = 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 described on the common chord of the circles x2 + y2 -8x + y -15 = 0 and x2 + y2 - 4x + 4y - 42 = 0 as diameter isa)x2 + y2- x + 2y + 4 = 0b)x2+ y2 + 10x - 2y + 12 = 0c)x2 + y2 - 12x - 2y + 12 = 0d)x2+ y2 - 5x + 3y + 7 = 0Correct answer is option 'C'. Can you explain this answer? tests, examples and also practice JEE tests.
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