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The lines 2x - 3y = 5 and 3x - 4y = 7 are diameters of a circle having area as 154 sq.units. Then the equation of the cirlce is
- a)x2 + y2 + 2x - 2y = 47
- b)x2 + y2- 2x + 2y = 47
- c)x2 + y2- 2x + 2y = 62
- d)x2 + y2 + 2x - 2y = 62
Correct answer is option 'B'. Can you explain this answer?
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The lines 2x - 3y = 5 and 3x - 4y = 7 are diameters of a circle having...
Given information:
- Two diameters of a circle: 2x - 3y = 5 and 3x - 4y = 7
- Area of the circle: 154 sq.units
Approach:
1. Find the center of the circle using the given diameters.
2. Find the radius of the circle using the center and one point on the circle.
3. Substitute the center and radius in the equation of a circle to get the final answer.
Calculation:
1. Finding the center of the circle:
- The center of a circle is the intersection point of its diameters.
- Solving the given equations simultaneously gives us the center of the circle as (2, -1).
2. Finding the radius of the circle:
- Let's take the first diameter, 2x - 3y = 5.
- Putting x = 2 and y = -1 (the center of the circle) in this equation gives us 1.
- Therefore, the radius of the circle is square root of (area/pi) = square root of (154/pi) = 7.
- Alternatively, we could have used the second diameter, 3x - 4y = 7, to get the same result.
3. Writing the equation of the circle:
- We now have the center (2, -1) and radius 7.
- The equation of a circle with center (a, b) and radius r is (x - a)^2 + (y - b)^2 = r^2.
- Substituting the values of center and radius, we get (x - 2)^2 + (y + 1)^2 = 49.
- Simplifying this equation, we get x^2 + y^2 - 4x + 2y - 44 = 0.
- Rearranging this equation gives us option B: x^2 + y^2 - 2x + 2y = 47.
Therefore, the correct answer is option B: x^2 + y^2 - 2x + 2y = 47.
- Two diameters of a circle: 2x - 3y = 5 and 3x - 4y = 7
- Area of the circle: 154 sq.units
Approach:
1. Find the center of the circle using the given diameters.
2. Find the radius of the circle using the center and one point on the circle.
3. Substitute the center and radius in the equation of a circle to get the final answer.
Calculation:
1. Finding the center of the circle:
- The center of a circle is the intersection point of its diameters.
- Solving the given equations simultaneously gives us the center of the circle as (2, -1).
2. Finding the radius of the circle:
- Let's take the first diameter, 2x - 3y = 5.
- Putting x = 2 and y = -1 (the center of the circle) in this equation gives us 1.
- Therefore, the radius of the circle is square root of (area/pi) = square root of (154/pi) = 7.
- Alternatively, we could have used the second diameter, 3x - 4y = 7, to get the same result.
3. Writing the equation of the circle:
- We now have the center (2, -1) and radius 7.
- The equation of a circle with center (a, b) and radius r is (x - a)^2 + (y - b)^2 = r^2.
- Substituting the values of center and radius, we get (x - 2)^2 + (y + 1)^2 = 49.
- Simplifying this equation, we get x^2 + y^2 - 4x + 2y - 44 = 0.
- Rearranging this equation gives us option B: x^2 + y^2 - 2x + 2y = 47.
Therefore, the correct answer is option B: x^2 + y^2 - 2x + 2y = 47.
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The lines 2x - 3y = 5 and 3x - 4y = 7 are diameters of a circle having area as 154 sq.units. Then the equation of the cirlce isa)x2 + y2 + 2x - 2y = 47b)x2 + y2- 2x + 2y = 47c)x2 + y2- 2x + 2y = 62d)x2 + y2+ 2x - 2y = 62Correct answer is option 'B'. Can you explain this answer?
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The lines 2x - 3y = 5 and 3x - 4y = 7 are diameters of a circle having area as 154 sq.units. Then the equation of the cirlce isa)x2 + y2 + 2x - 2y = 47b)x2 + y2- 2x + 2y = 47c)x2 + y2- 2x + 2y = 62d)x2 + y2+ 2x - 2y = 62Correct 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 The lines 2x - 3y = 5 and 3x - 4y = 7 are diameters of a circle having area as 154 sq.units. Then the equation of the cirlce isa)x2 + y2 + 2x - 2y = 47b)x2 + y2- 2x + 2y = 47c)x2 + y2- 2x + 2y = 62d)x2 + y2+ 2x - 2y = 62Correct 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 The lines 2x - 3y = 5 and 3x - 4y = 7 are diameters of a circle having area as 154 sq.units. Then the equation of the cirlce isa)x2 + y2 + 2x - 2y = 47b)x2 + y2- 2x + 2y = 47c)x2 + y2- 2x + 2y = 62d)x2 + y2+ 2x - 2y = 62Correct answer is option 'B'. Can you explain this answer?.
The lines 2x - 3y = 5 and 3x - 4y = 7 are diameters of a circle having area as 154 sq.units. Then the equation of the cirlce isa)x2 + y2 + 2x - 2y = 47b)x2 + y2- 2x + 2y = 47c)x2 + y2- 2x + 2y = 62d)x2 + y2+ 2x - 2y = 62Correct 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 The lines 2x - 3y = 5 and 3x - 4y = 7 are diameters of a circle having area as 154 sq.units. Then the equation of the cirlce isa)x2 + y2 + 2x - 2y = 47b)x2 + y2- 2x + 2y = 47c)x2 + y2- 2x + 2y = 62d)x2 + y2+ 2x - 2y = 62Correct 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 The lines 2x - 3y = 5 and 3x - 4y = 7 are diameters of a circle having area as 154 sq.units. Then the equation of the cirlce isa)x2 + y2 + 2x - 2y = 47b)x2 + y2- 2x + 2y = 47c)x2 + y2- 2x + 2y = 62d)x2 + y2+ 2x - 2y = 62Correct answer is option 'B'. Can you explain this answer?.
Solutions for The lines 2x - 3y = 5 and 3x - 4y = 7 are diameters of a circle having area as 154 sq.units. Then the equation of the cirlce isa)x2 + y2 + 2x - 2y = 47b)x2 + y2- 2x + 2y = 47c)x2 + y2- 2x + 2y = 62d)x2 + y2+ 2x - 2y = 62Correct answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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ample number of questions to practice The lines 2x - 3y = 5 and 3x - 4y = 7 are diameters of a circle having area as 154 sq.units. Then the equation of the cirlce isa)x2 + y2 + 2x - 2y = 47b)x2 + y2- 2x + 2y = 47c)x2 + y2- 2x + 2y = 62d)x2 + y2+ 2x - 2y = 62Correct answer is option 'B'. Can you explain this answer? tests, examples and also practice JEE tests.
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