`"If "(x-a)^(2)+(y-b)^(2)=c^(2),` for some `cgt0,` prove that `([1+((dy)/(dx))^(2)]^(3/2))/((d^(2)y)/(dx^(2)))` is a constant or independent of a and b.

Explanation

First we will calculate `dy/dx`

`(x-a)^2+(y-b)^2=c^2`

Differentiating w.r.t. x

`(d((x-a)^2+(y-b)^2))/dx=(d(c^2))/dx`

`(d(x-a)^2)/dx+(d(y-b)^2)/dx=0`

`2(x-a).(d(x-a))/dx+2(y-b).(d(y-b))/dx=0`

`2(x-a)(1-0)+2(y-b).(dy/dx-0)=0`

`2(x-a)+2(y-b).(dy/dx)=0`

`2(y-b).(dy/dx)=-2(x-a)`

`dy/dx=(-2(x-a))/(2(y-b))`

`dy/dx=(-(x-a))/(y-b)`

Again Differentiating w.r.t. x

`d/dx(dy/dx)=d/dx((-(x-a))/(y-b))`

`(d^2y)/(dx^2)=-d/dx((x-a)/(y-b))`

Using quotient rule

`(d^2y)/(dx^2)=-(((d(x-a))/dx(y-b)-(d(y-b))/dx(x-a))/(y-b)^2)`

`(d^2y)/(dx^2)=-(((y-b)-(-(x-a))/((y-b))(x-a))/(y-b)^2)`

`(d^2y)/(dx^2)=-(((y-b)^2+(x-a)^2)/((y-b)^2(y-b)))`

`(d^2y)/(dx^2)=-(-c)^2/((y-b)^3)`

Now, finding value of `([1+(dy/dx)^2]^(3/2))/((d^2y)/(dx^2))`

`([1+(dy/dx)^2]^(3/2))/((d^2y)/(dx^2))`

Putting values

`=([1+((-(x-a))/(y-b))^2]^(3/2))/((-c^2)/(y-b)^3)`

`=-([((y-b)^2+(x-a)^2)/(y-b)^2]^(3/2))/(c^2/(y-b)^3)`

`=-([(c^2)/(y-b)^2]^(3/2))/(c^2/(y-b)^3)`

`=-[(c^2)/(y-b)^2]^(3/2)xx(y-b)^3/c^2`

`=-((c)/(y-b))^(2xx3/2)xx(y-b)^3/c^2`

`=-((c)/(y-b))^(3)xx(y-b)^3/c^2`

`=-c^3/c^2xx(y-b)^3/(y-b)^(3)`

`=-c`

= constant

Which is constant independent of a & b

Hence Proved