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प्रश्न
The differential equation of all circles which passes through the origin and whose centre lies on the X-axis is ______
पर्याय
`(x^2 - y^2) - 2xydy/dx = 0`
`(x^2 - y^2) + 2xydy/dx = 0`
`(x^2 - y^2) - xydy/dx = 0`
`(x^2 - y^2) + xydy/dx = 0`
MCQ
रिकाम्या जागा भरा
उत्तर
The differential equation of all circles which passes through the origin and whose centre lies on the X-axis is `underline((x^2 - y^2) + 2xydy/dx = 0)`.
Explanation:
The system of circles which passes through origin and whose centre lies on the X-axis is
x2 + y2 - 2ax = 0 ................(i)
Differentiating w.r.t. x, we get
2x + 2y`dy/dx` - 2a = 0
⇒ `2a = 2x + 2ydy/dx` ......................(ii)
Substituting (ii) in (i), we get
`x^2 + y^2 - 2x^2 - 2xydy/dx = 0`
⇒ `(y^2 - x^2) - 2xydy/dx - 0`
⇒ `(x^2 - y^2) + 2xydy/dx = 0`
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Solution of a Differential Equation
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