Statement I: The differential equation corres | vedclass

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

Question

(C) Statement $I$: The equation of a circle with center $(0, \alpha)$ and radius $k$ is $x^2 + (y - \alpha)^2 = k^2$ ... $(i)$.Differentiating with re

Vedclass · Accepted Answer

Both Statement $I$ and Statement $II$ are true

Vedclass · Answer

(C) Statement $I$: The equation of a circle with center $(0, \alpha)$ and radius $k$ is $x^2 + (y - \alpha)^2 = k^2$ ... $(i)$.Differentiating with respect to $x$: $2x + 2(y - \alpha)\frac{dy}{dx} = 0 \Rightarrow y - \alpha = -x\frac{dx}{dy}$.Thus, $\alpha = y + x\frac{dx}{dy}$.Substituting $\alpha$ into $(i)$: $x^2 + (-x\frac{dx}{dy})^2 = k^2 \Rightarrow x^2 + x^2(\frac{dx}{dy})^2 = k^2$.$x^2(1 + (\frac{dx}{dy})^2) = k^2 \Rightarrow x^2(1 + \frac{1}{(dy/dx)^2}) = k^2 \Rightarrow x^2(\frac{(dy/dx)^2 + 1}{(dy/dx)^2}) = k^2$.$x^2(dy/dx)^2 + x^2 = k^2(dy/dx)^2 \Rightarrow (x^2 - k^2)(dy/dx)^2 + x^2 = 0$. Hence, Statement $I$ is true.Statement $II$: The circle passes through the origin and its center is on the $X$-axis, so center is $(\alpha, 0)$ and radius is $|\alpha|$.The equation is $(x - \alpha)^2 + y^2 = \alpha^2 \Rightarrow x^2 - 2x\alpha + \alpha^2 + y^2 = \alpha^2 \Rightarrow x^2 + y^2 = 2x\alpha \Rightarrow \alpha = \frac{x^2 + y^2}{2x}$.Differentiating $x^2 + y^2 = 2x\alpha$ with respect to $x$: $2x + 2y\frac{dy}{dx} = 2\alpha$.Substitute $\alpha$: $x + y\frac{dy}{dx} = \frac{x^2 + y^2}{2x} \Rightarrow 2x^2 + 2xy\frac{dy}{dx} = x^2 + y^2 \Rightarrow x^2 - y^2 + 2xy\frac{dy}{dx} = 0$. Hence, Statement $II$ is true.

Similar Questions

Which of the following differential equations has $y=x$ as one of its particular solutions?

The differential equation obtained by eliminating the arbitrary constants $a$ and $b$ from $xy = ae^x + be^{-x}$ is

Verify that the given function $x+y=\tan ^{-1} y$ is a solution of the differential equation $y^{2} y^{\prime}+y^{2}+1=0$.

If $l$ and $m$ are the degree and the order respectively of the differential equation of the family of all circles in the $XY$ plane with radius $5$ units,then $2l + 3m =$

The differential equation for which $y = ax^2 + bx + c$ is the general solution is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module