Suppose the solution of the differential equation [(d y)/(d x)=((2+) x- y+2)/( x-2 y-( -4 …

Question

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 :

Accepted Answer

$$\begin{aligned}
& \frac{d y}{d x}=\frac{(2+\alpha) x-\beta y+2}{\beta x-2 \alpha y-(\beta \gamma-4 \alpha)} \
& \beta x d y-2 \alpha y d y-(\beta \gamma-4 \alpha) d y \
& =2 x d x+\alpha x d x-\beta y d x+2 d x \
& \beta \int(x d y+y d y)-\alpha y^2-(\beta \gamma-4 x) y=x^2+\frac{\alpha x^2}{2}+2 x \
& \beta x y-\alpha y^2-(\beta \gamma-4 \alpha) y=x^2+\frac{\alpha x^2}{2}+2 x \
& \left(1+\frac{\alpha}{2}\right) x^2+\alpha y^2-\beta x y+2 x+(\beta \gamma-4 \alpha) y=0 \
& \because \text { this represents circle passing through origin } \
& \Rightarrow \beta=0 \text { and } 1+\frac{\alpha}{2}=\alpha \
& \Rightarrow \alpha=2
\end{aligned}$$

$$\begin{aligned}
& \therefore C: 2 x^2+2 y^2+2 x-8 y=0 \
& x^2+y^2+x-4 y=0 \
& \text { Radius }=\sqrt{\frac{1}{4}+4-0} \
& \quad=\frac{\sqrt{17}}{2}
\end{aligned}$$

Solutions · 1