Question

Form the differential equation by eliminating α and β from (x – α)² + (y – β)² = r²

Answer 1

(x – α)² + (y – α)² = γ²  .........(1)

Where α and ß are parameters.

Since equation (1) contains two orbitary constants,

We differentiate it two times w.r.t. x

Differentiating (1) w.r.t. x, we get

`2(x - alpha) + 2(y - beta) ("d"y)/("d"x)` = 0

⇒ `(x + alpha) + (y - beta) ("d"y)/("d"x)` = 0  ........(2)

Differentiating (2) w.r. to x, we get,

`1 + (y - beta) ("d"^2y)/("d"x^2) + (("d"y)/("d"x)) (("d"y)/("d"x))` = 0  ..........(3)

From (3), we have

`(y - beta) ("d"^2y)/("d"x^2) = - [1 + (("d"y)/("d"x))^2]`

`(y - beta) = - ([1 +(("d"y)/("d"x))^2])/((("d"^2y)/("d"x^2))`  ........(4)

Putting the value of `(y - beta)` in (2), we obtain,

`(x - alpha)  - ([1 + (("d"y)/("d"x))^2])/((("d"^2y)/("d"x^2))) [("d"y)/("d"x)]` = 0

`(x - alpha) = ([1 + (("d"y)/("d"x))^2])/((("d"^2y)/("d"x^2)))`  .......(5)

Substituting the value of (x – α) and (y – ß) in (5) we get

`({1 + (("d"y)/("d"x))^2} [(("d"y)/("d"x))]^2)/(("d"^2y)/("d"x^2))^2 + ({1 + (("d"y)/("d"x))^2}^2)/(("d"^2y)/("d"x^2))^2 = "r"^2`

`({1 + (("d"y)/("d"x))^2} [("d"y)/("d"x)]^2 + {1 + (("d"y)/("d"x))^2}^2)/(("d"^2y)/("d"x^2))^2 = "r"^2`

`({1 + (("d"y)/("d"x))^2}^2 + {(("d"y)/("d"x))^2 + 1})/(("d"^2y)/("d"x^2))^2 = "r"^2`

`{1 + (("d"y)/("d"x))^2} ="r"^2 (("d"^2y)/("d"x^2))^2`

This is the required differential equation.