Differential Equation: The differential equation of all circles passing through the origin… | AIEEE 2007

AIEEE 2007 (Offline)Question 1

The differential equation of all circles passing through the origin and having their centres on the $x$-axis is :

Differential Equation

$y^{2}=x^{2}+2xy\frac{dy}{dx}$✓ Correct answer$y^{2}=x^{2}-2xy\frac{dy}{dx}$$x^{2}=y^{2}+xy\frac{dy}{dx}$$x^{2}=y^{2}+3xy\frac{dy}{dx}$

Solution

Concept. Circles through the origin with centre $(a,0)$: $x^2+y^2=2ax$; one parameter. Formulas used. (1) $x^2+y^2=2ax\Rightarrow2x+2yy'=2a$Step 1 - Find $2a$. From the circle, $2a=\dfrac{x^2+y^2}{x}$. Step 2 - Substitute in (1). $$2x+2yy'=\frac{x^2+y^2}{x}\ \Longrightarrow\ 2x^2+2xyy'=x^2+y^2.$$ Step 3 - Rearrange. $y^2=x^2+2xy\,\dfrac{dy}{dx}$. Answer: (A)

Differential Equation - JEE Mathematics question (AIEEE 2007 (Offline))

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