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A circle passes through the origin and point (5/2,1/2)and has. 2y-3x=0as a diameter. Find its equation.?
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A circle passes through the origin and point (5/2,1/2)and has. 2y-3x=0...
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A circle passes through the origin and point (5/2,1/2)and has. 2y-3x=0...
Problem: Find the equation of a circle that passes through the origin and point (5/2,1/2) and has 2y-3x=0 as a diameter.
Solution:
Step 1: Finding the center of the circle
The diameter of the circle is 2y-3x=0, which can be written as y=(3/2)x. This is the equation of a line passing through the center of the circle. To find the center, we need to find the midpoint of the diameter.
The midpoint of the diameter is ((0+5/2)/2, (0+1/2)/2) = (5/4, 1/4).
Therefore, the center of the circle is (5/4, 1/4).
Step 2: Finding the radius of the circle
The radius of the circle is the distance between the center and any point on the circle. We can use the distance formula to find the radius.
Let (x,y) be a point on the circle. Then, by the distance formula,
sqrt((x-5/4)^2 + (y-1/4)^2) = sqrt((5/4-0)^2 + (1/4-0)^2)
Simplifying, we get:
(x-5/4)^2 + (y-1/4)^2 = 1
Therefore, the equation of the circle is:
(x-5/4)^2 + (y-1/4)^2 = 1
Conclusion: The equation of the circle that passes through the origin and point (5/2,1/2) and has 2y-3x=0 as a diameter is (x-5/4)^2 + (y-1/4)^2 = 1.
Solution:
Step 1: Finding the center of the circle
The diameter of the circle is 2y-3x=0, which can be written as y=(3/2)x. This is the equation of a line passing through the center of the circle. To find the center, we need to find the midpoint of the diameter.
The midpoint of the diameter is ((0+5/2)/2, (0+1/2)/2) = (5/4, 1/4).
Therefore, the center of the circle is (5/4, 1/4).
Step 2: Finding the radius of the circle
The radius of the circle is the distance between the center and any point on the circle. We can use the distance formula to find the radius.
Let (x,y) be a point on the circle. Then, by the distance formula,
sqrt((x-5/4)^2 + (y-1/4)^2) = sqrt((5/4-0)^2 + (1/4-0)^2)
Simplifying, we get:
(x-5/4)^2 + (y-1/4)^2 = 1
Therefore, the equation of the circle is:
(x-5/4)^2 + (y-1/4)^2 = 1
Conclusion: The equation of the circle that passes through the origin and point (5/2,1/2) and has 2y-3x=0 as a diameter is (x-5/4)^2 + (y-1/4)^2 = 1.
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A circle passes through the origin and point (5/2,1/2)and has. 2y-3x=0as a diameter. Find its equation.?
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A circle passes through the origin and point (5/2,1/2)and has. 2y-3x=0as a diameter. Find its equation.? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A circle passes through the origin and point (5/2,1/2)and has. 2y-3x=0as a diameter. Find its equation.? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A circle passes through the origin and point (5/2,1/2)and has. 2y-3x=0as a diameter. Find its equation.?.
A circle passes through the origin and point (5/2,1/2)and has. 2y-3x=0as a diameter. Find its equation.? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A circle passes through the origin and point (5/2,1/2)and has. 2y-3x=0as a diameter. Find its equation.? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A circle passes through the origin and point (5/2,1/2)and has. 2y-3x=0as a diameter. Find its equation.?.
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