The differential equation of the family of circles passing through the origin and having centre at the line y=x is :

Differential Equations PYQ — JEE Main 2024 · 5 Apr · Shift 2 · Q50 with Solution

View written solutionFree

Correct answer: D

Equation of the family of circles

Since the centre lies on the line $y=x$ , let the centre be $(a,a)$ .

Also, the circle passes through the origin $(0,0)$ , so its radius is $r=\sqrt{a^2+a^2}=\sqrt{2}a.$

Hence the family of circles is

(x-a)^2+(y-a)^2=2a^2.

Expanding,

x^2-2ax+a^2+y^2-2ay+a^2=2a^2

which simplifies to

x^2+y^2-2a(x+y)=0.

So the family is

x^2+y^2=2a(x+y). \qquad (1)

Differentiate with respect to $x$

Treating $y$ as a function of $x$ and $a$ as parameter,

\frac{d}{dx}(x^2+y^2)=\frac{d}{dx}[2a(x+y)].

So,

2x+2y\frac{dy}{dx}=2a\left(1+\frac{dy}{dx}\right).

Thus,

x+y\frac{dy}{dx}=a\left(1+\frac{dy}{dx}\right). \qquad (2)

From (2),

a=\frac{x+y\frac{dy}{dx}}{1+\frac{dy}{dx}}. \qquad (3)

Eliminate the parameter $a$

From (1),

a=\frac{x^2+y^2}{2(x+y)}. \qquad (4)

Equating (3) and (4),

\frac{x+y\frac{dy}{dx}}{1+\frac{dy}{dx}}=\frac{x^2+y^2}{2(x+y)}.

Cross-multiplying,

2(x+y)\left(x+y\frac{dy}{dx}\right)=(x^2+y^2)\left(1+\frac{dy}{dx}\right).

Expand both sides:

2x(x+y)+2y(x+y)\frac{dy}{dx}=x^2+y^2+(x^2+y^2)\frac{dy}{dx}.

That is,

2x^2+2xy+\left(2xy+2y^2\right)\frac{dy}{dx}=x^2+y^2+(x^2+y^2)\frac{dy}{dx}.

Rearranging,

(x^2+2xy-y^2)+(-x^2+2xy+y^2)\frac{dy}{dx}=0.

So,

(x^2-y^2+2xy)+( -x^2+y^2+2xy)\frac{dy}{dx}=0.

Hence,

(x^2-y^2+2xy)=(x^2-y^2-2xy)\frac{dy}{dx}.

Therefore,

(x^2-y^2+2xy)\,dx=(x^2-y^2-2xy)\,dy.

Match with options

This matches Option D:

\left(x^2-y^2+2xy\right)dx=\left(x^2-y^2-2xy\right)dy.

More from Differential Equations

View written solutionFree

Correct answer: D

Equation of the family of circles

Since the centre lies on the line $y=x$ , let the centre be $(a,a)$ .

Also, the circle passes through the origin $(0,0)$ , so its radius is $r=\sqrt{a^2+a^2}=\sqrt{2}a.$

Hence the family of circles is

(x-a)^2+(y-a)^2=2a^2.

Expanding,

x^2-2ax+a^2+y^2-2ay+a^2=2a^2

which simplifies to

x^2+y^2-2a(x+y)=0.

So the family is

x^2+y^2=2a(x+y). \qquad (1)

Differentiate with respect to $x$

Treating $y$ as a function of $x$ and $a$ as parameter,

\frac{d}{dx}(x^2+y^2)=\frac{d}{dx}[2a(x+y)].

So,

2x+2y\frac{dy}{dx}=2a\left(1+\frac{dy}{dx}\right).

Thus,

x+y\frac{dy}{dx}=a\left(1+\frac{dy}{dx}\right). \qquad (2)

From (2),

a=\frac{x+y\frac{dy}{dx}}{1+\frac{dy}{dx}}. \qquad (3)

Eliminate the parameter $a$

From (1),

a=\frac{x^2+y^2}{2(x+y)}. \qquad (4)

Equating (3) and (4),

\frac{x+y\frac{dy}{dx}}{1+\frac{dy}{dx}}=\frac{x^2+y^2}{2(x+y)}.

Cross-multiplying,

2(x+y)\left(x+y\frac{dy}{dx}\right)=(x^2+y^2)\left(1+\frac{dy}{dx}\right).

Expand both sides:

2x(x+y)+2y(x+y)\frac{dy}{dx}=x^2+y^2+(x^2+y^2)\frac{dy}{dx}.

That is,

2x^2+2xy+\left(2xy+2y^2\right)\frac{dy}{dx}=x^2+y^2+(x^2+y^2)\frac{dy}{dx}.

Rearranging,

(x^2+2xy-y^2)+(-x^2+2xy+y^2)\frac{dy}{dx}=0.

So,

(x^2-y^2+2xy)+( -x^2+y^2+2xy)\frac{dy}{dx}=0.

Hence,

(x^2-y^2+2xy)=(x^2-y^2-2xy)\frac{dy}{dx}.

Therefore,

(x^2-y^2+2xy)\,dx=(x^2-y^2-2xy)\,dy.

Match with options

This matches Option D:

\left(x^2-y^2+2xy\right)dx=\left(x^2-y^2-2xy\right)dy.

Differential Equations question

Differential Equations question

More from Differential Equations

Differential Equations question

Differential Equations question

More from Differential Equations