Suppose the solution of the differential equation (d y)/(d x)=((2+α) x-β y+2)/(β x-2 α y-(β γ-4 α)) represents a circle passing through origin. Then the radius of this circle is :

Correct answer: D. 17 2 aligned & (d y)/(d x)=((2+α) x-β y+2)/(β x-y(2 α+β)+4 α) & β x d y-(2 α+β) y d y+4 α d y=(2+α) x d x-β y d x+2 d x & β(x d y+y d x)-(2 α+β) y d y+4 α d y=(2+α) x d x+2 d x & β x y-((2 α+β) y^2)/(2)+4 α y=((2+α) x^2)/(2) aligned aligned & β=0 for this to be circle & (2+α) (x^2)/(2)+α y^2+2 x-4 α y=0 aligned aligned & i.e. 2 x^2+2 y^2+2 x-8 y=0 & x^2+y^2+x-4 y=0 & r d= (1)/(4)+4 = 17 2 aligned

JEE Main 20246 AprilEvening Shift

Exam: JEE Main
Year: 2024
Date: 6 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Differential Equations

JEE Main 2024 · 6 April · Evening Shift · Mathematics · Differential Equations

Suppose the solution of the differential equation (d y)/(d...

Suppose the solution of the differential equation

\frac{d y}{d x}=\frac{(2+\alpha) x-\beta y+2}{\beta x-2 \alpha y-(\beta \gamma-4 \alpha)}

represents a circle passing through origin. Then the radius of this circle is :

Options

A2
B $\sqrt{17}$
C $\frac{1}{2}$
D $\frac{\sqrt{17}}{2}$

Correct answer

D. $\frac{\sqrt{17}}{2}$

Step-by-step solution

\begin{aligned} & \frac{d y}{d x}=\frac{(2+\alpha) x-\beta y+2}{\beta x-y(2 \alpha+\beta)+4 \alpha} \\ & \beta x d y-(2 \alpha+\beta) y d y+4 \alpha d y=(2+\alpha) x d x-\beta y d x+2 d x \\ & \beta(x d y+y d x)-(2 \alpha+\beta) y d y+4 \alpha d y=(2+\alpha) x d x+2 d x \\ & \beta x y-\frac{(2 \alpha+\beta) y^2}{2}+4 \alpha y=\frac{(2+\alpha) x^2}{2}\end{aligned}

\begin{aligned} & \Rightarrow \beta=0 \text{ for this to be circle } \\ & (2+\alpha) \frac{x^2}{2}+\alpha y^2+2 x-4 \alpha y=0\end{aligned}

\begin{aligned} & \text{ i.e. } 2 x^2+2 y^2+2 x-8 y=0 \\ & x^2+y^2+x-4 y=0 \\ & r d=\sqrt{\frac{1}{4}+4}=\frac{\sqrt{17}}{2}\end{aligned}

Practice Differential Equations on Quantrex Academy →

More from Differential Equations