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The equation of the circle passing through intersection of circles x2+y2-1=0 and x2+y2-2x-4y+1=0 and touching x+2y=0 is
- a)x2+y2-x-2y=0
- b)x2+y2-x+2y=0
- c)x2+y2+x-2y=0
- d)x2+y2+x+2y=0
Correct answer is option 'A'. Can you explain this answer?
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The equation of the circle passing through intersection of circles x2+...
Solution:
The given equations of circles are,
x2 + y2 = 1 ...(1)
x2 + y2 - 2x - 4y + 1 = 0 ...(2)
Intersection of the 2 circles (1) and (2) can be found by solving them simultaneously.
Subtracting equation (1) from equation (2), we get
-2x - 4y + 1 - 1 = 0 - 1
-2x - 4y = -1
2x + 4y = 1 ...(3)
Multiplying equation (1) by 4, we get
4x2 + 4y2 = 4 ...(4)
Adding equations (3) and (4), we get
4x2 + 2x + 4y2 - 4y + 1 = 4 + 1
4x2 + 2x + 4y2 - 4y - 3 = 0
x2 + y2 + (2x - 4y)/2 - 3/4 = 0
x2 + y2 + x - 2y - 3/4 = 0 ...(5)
Equation (5) represents the required circle.
Now, we need to check whether this circle touches the line x = 2y or not.
For a circle to touch a line, the distance between their centers should be equal to the radius of the circle.
The center of the circle (5) can be found by differentiating the equation and equating it to zero.
2x + 1 = 0
x = -1/2
2y - 2 = 0
y = 1
Hence, the center of the circle is (-1/2, 1).
The radius of the circle can be found by substituting the center coordinates in the equation of the circle.
(-1/2)2 + 12 - (-1/2) - 2(1) - 3/4 = 0
1/4 + 1 - 1/2 - 2 - 3/4 = 0
-5/4 = 0
Since the radius is negative, the circle cannot touch the line x = 2y.
Therefore, the given problem statement is incorrect and there is no correct option among the given choices.
The given equations of circles are,
x2 + y2 = 1 ...(1)
x2 + y2 - 2x - 4y + 1 = 0 ...(2)
Intersection of the 2 circles (1) and (2) can be found by solving them simultaneously.
Subtracting equation (1) from equation (2), we get
-2x - 4y + 1 - 1 = 0 - 1
-2x - 4y = -1
2x + 4y = 1 ...(3)
Multiplying equation (1) by 4, we get
4x2 + 4y2 = 4 ...(4)
Adding equations (3) and (4), we get
4x2 + 2x + 4y2 - 4y + 1 = 4 + 1
4x2 + 2x + 4y2 - 4y - 3 = 0
x2 + y2 + (2x - 4y)/2 - 3/4 = 0
x2 + y2 + x - 2y - 3/4 = 0 ...(5)
Equation (5) represents the required circle.
Now, we need to check whether this circle touches the line x = 2y or not.
For a circle to touch a line, the distance between their centers should be equal to the radius of the circle.
The center of the circle (5) can be found by differentiating the equation and equating it to zero.
2x + 1 = 0
x = -1/2
2y - 2 = 0
y = 1
Hence, the center of the circle is (-1/2, 1).
The radius of the circle can be found by substituting the center coordinates in the equation of the circle.
(-1/2)2 + 12 - (-1/2) - 2(1) - 3/4 = 0
1/4 + 1 - 1/2 - 2 - 3/4 = 0
-5/4 = 0
Since the radius is negative, the circle cannot touch the line x = 2y.
Therefore, the given problem statement is incorrect and there is no correct option among the given choices.
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The equation of the circle passing through intersection of circles x2+y2-1=0 and x2+y2-2x-4y+1=0 and touching x+2y=0 isa)x2+y2-x-2y=0b)x2+y2-x+2y=0c)x2+y2+x-2y=0d)x2+y2+x+2y=0Correct answer is option 'A'. Can you explain this answer?
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The equation of the circle passing through intersection of circles x2+y2-1=0 and x2+y2-2x-4y+1=0 and touching x+2y=0 isa)x2+y2-x-2y=0b)x2+y2-x+2y=0c)x2+y2+x-2y=0d)x2+y2+x+2y=0Correct answer is option 'A'. 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 passing through intersection of circles x2+y2-1=0 and x2+y2-2x-4y+1=0 and touching x+2y=0 isa)x2+y2-x-2y=0b)x2+y2-x+2y=0c)x2+y2+x-2y=0d)x2+y2+x+2y=0Correct answer is option 'A'. 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 passing through intersection of circles x2+y2-1=0 and x2+y2-2x-4y+1=0 and touching x+2y=0 isa)x2+y2-x-2y=0b)x2+y2-x+2y=0c)x2+y2+x-2y=0d)x2+y2+x+2y=0Correct answer is option 'A'. Can you explain this answer?.
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