The Differential Equation representing the family of tangents to the circle $x^2+y^2 =r^2$, is :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → S$\frac{d^2y}{dx^2}=0$
$x^2+y^2 =r^2$
so differentiating wrt x
we get $2x+2y\frac{dy}{dx}=0$
$\frac{dy}{dx}=-frac{x}{y}$
let point of tangent be (a, b)
eq:- $y-b=-\frac{a}{b}(x-a)$
$by-b^2=-ax+a^2$
$ax+by=a^2+b^2$
differentiating wrt x
$a+b\frac{dy}{dx}=0$
so $\frac{dy}{dx}=-\frac{a}{b}$
$⇒\frac{d^2y}{dx^2}=0$