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प्रश्न
Form the differential equation of all circles which pass through origin and whose centres lie on y-axis.
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उत्तर
Equation of circle which passes through the origin and whose centre lies on y-axis is
(x – 0)2 + (y – a)2 = a2
⇒ x2 + y2 + a2 – 2ay = a2
⇒ x2 + y2 – 2ay = 0 ......(i)
Differentiating both sides w.r.t. x we get
⇒ `2x + 2y * "dy"/"dx" - 2"a" * "dy"/"dx"` = 0
⇒ `x + y "dy"/"dx" - "a" * "dy"/"dx"` = 0
⇒ `x + (y - "a") * "dy"/"dx"` = 0
`y - "a" = x/("dy"/"dx")`
a = `y + (-x)/("dy"/"dx")`
Putting the value of a in equation (i), we get
`x^2 + y^2 - 2(y + x/("dy"/"dx"))y` = 0
⇒ `x^2 + y^2 - 2y^2 - (2xy)/("dy"/"dx")` = 0
⇒ `x^2 - y^2 = (2xy)/("dy"/"dx")`
∴ `(x^2 - y^2) "dy"/"dx" - 2xy` = 0
Hence, the required differential equation is `(x^2 - y^2) "dy"/"dx" - 2xy` = 0
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