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प्रश्न
Find the differential equation of all circles having radius 9 and centre at point (h, k).
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उत्तर
Equation of the circle having radius 9 and centre at point (h, k) is
(x - h)2 + (y - k)2 = 81, .....(1)
where h and k are arbitrary constant.
Differentiating (1) w.r.t. x, we get
`2("x - h") * "d"/"dx" ("x - h") + 2 ("y - k") * "d"/"dx" ("y - k") = 0`
∴ (x - h)(1 - 0) + (y - k)`("dy"/"dx" - 0) = 0`
∴ (x - h) + (y - k) `"dy"/"dx" = 0` .....(2)
Differentiating again w.r.t. x, we get
`"d"/"dx" ("x - h") + ("y - k") * "d"/"dx"("dy"/"dx") + "dy"/"dx" * "d"/"dx" ("y - k") = 0`
∴ `(1 - 0) + ("y - k") ("d"^2"y")/"dx"^2 + "dy"/"dx" * ("dy"/"dx" - 0) = 0`
∴ `("y - k") ("d"^2"y")/"dx"^2 + ("dy"/"dx")^2` + 1 = 0
∴ `("y - k") ("d"^2"y")/"dx"^2 = - [("dy"/"dx")^2 + 1]`
∴ `"y - k" = (- ("dy"/"dx")^2 + 1)/(("d"^2"y")/"dx"^2` ....(3)
From (2), x - h = - (y - k)`"dy"/"dx"`
Substituting the value of (x - h) in (1), we get
`("y - k")^2 ("dy"/"dx")^2 + ("y - k")^2 = 81`
∴ `("dy"/"dx")^2 + 1 = 81/("y - k")^2`
∴ `("dy"/"dx")^2 + 1 = (81 * ("d"^2"y")/"dx"^2)/[("dy"/"dx")^2 + 1]^2`
∴ `81 (("d"^2"y")/"dx"^2)^2 = [("dy"/"dx")^2 + 1]^3`
This is the required D.E.
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