Maharashtra State BoardHSC Arts 12th Board Exam
Advertisement Remove all ads

Find the differential equation of all circles having radius 9 and centre at point (h, k). - Mathematics and Statistics

Sum

Find the differential equation of all circles having radius 9 and centre at point (h, k).

Advertisement Remove all ads

Solution

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.

Concept: Formation of Differential Equations
  Is there an error in this question or solution?
Advertisement Remove all ads

APPEARS IN

Advertisement Remove all ads
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×