Given that the slope of the tangent to a curv | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The equation of the given circle is $x^2+y^2-2x-2y=0$. Completing the square,we get $(x-1)^2+(y-1)^2=2$. Thus,the centre of the circle is $(1, 1)$

Vedclass · Accepted Answer

$x \log |y|=2(x-1)$

Vedclass · Answer

(C) The equation of the given circle is $x^2+y^2-2x-2y=0$. Completing the square,we get $(x-1)^2+(y-1)^2=2$. Thus,the centre of the circle is $(1, 1)$. The slope of the tangent to the curve is given by $\frac{dy}{dx} = \frac{2y}{x^2}$. Separating the variables,we have $\int \frac{1}{y} dy = \int \frac{2}{x^2} dx$. Integrating both sides,we get $\log |y| = -\frac{2}{x} + c$. Since the curve passes through $(1, 1)$,we substitute $x=1$ and $y=1$: $\log |1| = -\frac{2}{1} + c \implies 0 = -2 + c \implies c = 2$. Substituting $c=2$ into the general solution,we get $\log |y| = -\frac{2}{x} + 2$. Multiplying by $x$,we get $x \log |y| = -2 + 2x = 2(x-1)$.

Given that the slope of the tangent to a curve $y=y(x)$ at any point $(x, y)$ is $\frac{2y}{x^2}$. If the curve passes through the centre of the circle $x^2+y^2-2x-2y=0$,then its equation is

Similar Questions

The general solution of the differential equation $e^{\frac{1}{2}\left(\frac{dy}{dx}\right)}=3^x$ is (where $C$ is a constant of integration).

The solution of the differential equation $\log \left(\frac{dy}{dx}\right) = 9x - 6y + 6$ is (given that $y = 1$ when $x = 0$):

On solving $\frac{dy}{dx} = \frac{x-y+3}{2x-2y+5}$,the solution obtained is $x = 2(x-y) + \log(t) + c$,find $t$.

The solution of $\tan y \frac{dy}{dx} = \sin(x+y) + \sin(x-y)$ is

$A$ differentiable function satisfies $3f^2(x) f'(x) = 2x$. Given $f(2) = 1$,then the value of $f(3)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module