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The differential equation of all circles which pass through (0,0) and having centre on y-axis, is
- a)(x2-y2)(dy/dx)-2xy=0
- b)(x2-y2))(dy/dx)+2xy=0
- c)(x2-y2)(dy/dx)-xy=0
- d)(x2-y2))(dy/dx)+xy=0
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
The differential equation of all circles which pass through (0,0) and ...
Explanation:
Given: Circle passes through (0,0) and has center on y-axis.
Given: Circle passes through (0,0) and has center on y-axis.
- Equation of a circle: The general equation of a circle with center at (h,k) and radius r is given by: (x-h)^2 + (y-k)^2 = r^2.
- Circle passing through (0,0): Since the circle passes through (0,0), we have: (0-h)^2 + (0-k)^2 = r^2. This simplifies to h^2 + k^2 = r^2.
- Center on y-axis: Since the center lies on the y-axis, the x-coordinate of the center (h) is 0.
- Substitute h=0: Substituting h=0 in the equation h^2 + k^2 = r^2, we get k^2 = r^2.
- Equation of circle: Therefore, the equation of the circle becomes: x^2 + (y-k)^2 = k^2.
- Expand and simplify: Expanding and simplifying the equation, we get: x^2 + y^2 - 2ky = 0.
- Equation in terms of dy/dx: To express the equation in terms of dy/dx, we differentiate both sides with respect to x.
- Final differential equation: After differentiation, we get: (x^2 - y^2) * dy/dx - 2xy = 0. Hence, the correct option is (a).
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The differential equation of all circles which pass through (0,0) and having centre on y-axis, isa)(x2-y2)(dy/dx)-2xy=0b)(x2-y2))(dy/dx)+2xy=0c)(x2-y2)(dy/dx)-xy=0d)(x2-y2))(dy/dx)+xy=0Correct answer is option 'A'. Can you explain this answer?
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