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A circle having area = 154 sq. units has two diameters 2x–3y–5 = 0 and 3x–4y–7 0, then the equation of the circle is
  • a)
    x2+y2+2x+2y−47=0
  • b)
    x2+y2−2x−2y−47=0
  • c)
    x2+y2−2x+2y−47=0
  • d)
    x2+y2+2x−2y−47=0
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
A circle having area = 154 sq. units has two diameters2x–3y&ndas...
You just need to find the coordinates of center of the circle, which can be found by finding the point of intersection of both the diameters and then by using the formula for the area of a circle, you can easily calculate the value of radius of the circle, and then write down the equation. :)
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A circle having area = 154 sq. units has two diameters2x–3y–5 = 0 and 3x–4y–7 0,then the equation of the circle isa)x2+y2+2x+2y−47=0b)x2+y2−2x−2y−47=0c)x2+y2−2x+2y−47=0d)x2+y2+2x−2y−47=0Correct answer is option 'C'. Can you explain this answer?
Question Description
A circle having area = 154 sq. units has two diameters2x–3y–5 = 0 and 3x–4y–7 0,then the equation of the circle isa)x2+y2+2x+2y−47=0b)x2+y2−2x−2y−47=0c)x2+y2−2x+2y−47=0d)x2+y2+2x−2y−47=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 A circle having area = 154 sq. units has two diameters2x–3y–5 = 0 and 3x–4y–7 0,then the equation of the circle isa)x2+y2+2x+2y−47=0b)x2+y2−2x−2y−47=0c)x2+y2−2x+2y−47=0d)x2+y2+2x−2y−47=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 A circle having area = 154 sq. units has two diameters2x–3y–5 = 0 and 3x–4y–7 0,then the equation of the circle isa)x2+y2+2x+2y−47=0b)x2+y2−2x−2y−47=0c)x2+y2−2x+2y−47=0d)x2+y2+2x−2y−47=0Correct answer is option 'C'. Can you explain this answer?.
Solutions for A circle having area = 154 sq. units has two diameters2x–3y–5 = 0 and 3x–4y–7 0,then the equation of the circle isa)x2+y2+2x+2y−47=0b)x2+y2−2x−2y−47=0c)x2+y2−2x+2y−47=0d)x2+y2+2x−2y−47=0Correct answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
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