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Equation of the circle having diameters 2x - 3y = 5 and 3x - 4y = 7 and radius 8 is
- a)x2 + y2 - 2x + 2y - 62 = 0
- b)x2 + y2 + 2x + 2y - 2 = 0
- c)x2 + y2 + 2x - 2y - 62 = 0
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
Equation of the circle having diameters 2x - 3y = 5 and 3x - 4y = 7 an...
Finding the Center of the Circle
To find the equation of the circle, we first need to find its center. We can do this by finding the intersection of the two diameters.
- Rewrite both equations in slope-intercept form:
2x - 3y = 5 --> y = (2/3)x - 5/3
3x - 4y = 7 --> y = (3/4)x - 7/4
- Find the midpoint of each diameter. This is the center of the circle.
Midpoint of first diameter:
( (0 + 5)/2 , (0 + 0)/2 ) = (5/2, 0)
Midpoint of second diameter:
( (0 + 7)/2 , (0 + 0)/2 ) = (7/2, 0)
- Equate the x-coordinates of the two midpoints and solve for y.
5/2 = 7/2 + m --> m = -1
- The center of the circle is at (5/2, -1).
Finding the Equation of the Circle
Now that we have the center of the circle, we can use the radius of 8 to write the equation in standard form.
- The standard form of the equation of a circle is (x - h)^2 + (y - k)^2 = r^2, where (h,k) is the center of the circle and r is the radius.
- Substituting (5/2, -1) for (h,k) and 8 for r, we get:
(x - 5/2)^2 + (y + 1)^2 = 64
- Expanding and simplifying, we get:
x^2 - 5x + 25/4 + y^2 + 2y + 1 = 64
x^2 + y^2 - 5x + 2y - 62 = 0
- Therefore, the equation of the circle is x^2 + y^2 - 5x + 2y - 62 = 0, which is option A.
To find the equation of the circle, we first need to find its center. We can do this by finding the intersection of the two diameters.
- Rewrite both equations in slope-intercept form:
2x - 3y = 5 --> y = (2/3)x - 5/3
3x - 4y = 7 --> y = (3/4)x - 7/4
- Find the midpoint of each diameter. This is the center of the circle.
Midpoint of first diameter:
( (0 + 5)/2 , (0 + 0)/2 ) = (5/2, 0)
Midpoint of second diameter:
( (0 + 7)/2 , (0 + 0)/2 ) = (7/2, 0)
- Equate the x-coordinates of the two midpoints and solve for y.
5/2 = 7/2 + m --> m = -1
- The center of the circle is at (5/2, -1).
Finding the Equation of the Circle
Now that we have the center of the circle, we can use the radius of 8 to write the equation in standard form.
- The standard form of the equation of a circle is (x - h)^2 + (y - k)^2 = r^2, where (h,k) is the center of the circle and r is the radius.
- Substituting (5/2, -1) for (h,k) and 8 for r, we get:
(x - 5/2)^2 + (y + 1)^2 = 64
- Expanding and simplifying, we get:
x^2 - 5x + 25/4 + y^2 + 2y + 1 = 64
x^2 + y^2 - 5x + 2y - 62 = 0
- Therefore, the equation of the circle is x^2 + y^2 - 5x + 2y - 62 = 0, which is option A.
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Community Answer
Equation of the circle having diameters 2x - 3y = 5 and 3x - 4y = 7 an...
The intersection of the diagonals i.e. (1,-1) is the centre of the circle with radius 8. put in the formula
(x-1)^2+(y+1)^2=8^2
x2+y2-2x+2y-62=0
(x-1)^2+(y+1)^2=8^2
x2+y2-2x+2y-62=0
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Equation of the circle having diameters 2x - 3y = 5 and 3x - 4y = 7 and radius 8 isa)x2 + y2 - 2x + 2y - 62 = 0b)x2 + y2 + 2x + 2y - 2 = 0c)x2 + y2 + 2x - 2y - 62 = 0d)none of theseCorrect answer is option 'A'. 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 Equation of the circle having diameters 2x - 3y = 5 and 3x - 4y = 7 and radius 8 isa)x2 + y2 - 2x + 2y - 62 = 0b)x2 + y2 + 2x + 2y - 2 = 0c)x2 + y2 + 2x - 2y - 62 = 0d)none of theseCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Equation of the circle having diameters 2x - 3y = 5 and 3x - 4y = 7 and radius 8 isa)x2 + y2 - 2x + 2y - 62 = 0b)x2 + y2 + 2x + 2y - 2 = 0c)x2 + y2 + 2x - 2y - 62 = 0d)none of theseCorrect answer is option 'A'. Can you explain this answer?.
Equation of the circle having diameters 2x - 3y = 5 and 3x - 4y = 7 and radius 8 isa)x2 + y2 - 2x + 2y - 62 = 0b)x2 + y2 + 2x + 2y - 2 = 0c)x2 + y2 + 2x - 2y - 62 = 0d)none of theseCorrect answer is option 'A'. 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 Equation of the circle having diameters 2x - 3y = 5 and 3x - 4y = 7 and radius 8 isa)x2 + y2 - 2x + 2y - 62 = 0b)x2 + y2 + 2x + 2y - 2 = 0c)x2 + y2 + 2x - 2y - 62 = 0d)none of theseCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Equation of the circle having diameters 2x - 3y = 5 and 3x - 4y = 7 and radius 8 isa)x2 + y2 - 2x + 2y - 62 = 0b)x2 + y2 + 2x + 2y - 2 = 0c)x2 + y2 + 2x - 2y - 62 = 0d)none of theseCorrect answer is option 'A'. Can you explain this answer?.
Solutions for Equation of the circle having diameters 2x - 3y = 5 and 3x - 4y = 7 and radius 8 isa)x2 + y2 - 2x + 2y - 62 = 0b)x2 + y2 + 2x + 2y - 2 = 0c)x2 + y2 + 2x - 2y - 62 = 0d)none of theseCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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