Question

find dy/dx if x and y are related as (x-a)^2+ (y-b)^2=r^2​

Answer 1

Answer:

$\frac{dy}{dx} = -\frac{(x-a)}{(y-b)}$

Step-by-step explanation:

Given, $(x-a)^{2}+(y-b)^{2} = r^{2}$

⇒ $x^{2}+a^{2}-2ax+y^{2}+b^{2}-2by= r^{2}$

or $x^{2}-2ax+y^{2}-2by= r^{2}-a^{2}-b^{2}$

On differentiating both sides w.r.t. x, we get

⇒ $2x-2a+2y\frac{dy}{dx}-2b\frac{dy}{dx} = 0$

or $x-a+\frac{dy}{dx}(y-b) = 0$

or $\frac{dy}{dx} = -\frac{(x-a)}{(y-b)}$