JEE Mathematics 2024 Question with Solution

Given: A family of circles passes through the origin and has its center on the line $y=x$.

Find: The differential equation of this family.

The general equation for a circle with the center on $y=x$ passing through the origin is:

$$(x-h)^2+(y-h)^2=r^2$$

where $h$ is the parameter.

Since the circle passes through the origin, substitute $x=0$ and $y=0$:

$$h^2+h^2=r^2$$ $$2h^2=r^2$$

So the family becomes

$$(x-h)^2+(y-h)^2=2h^2$$

Expanding,

$$x^2-2hx+h^2+y^2-2hy+h^2=2h^2$$ $$x^2+y^2-2h(x+y)=0$$

Hence,

$$h=\frac{x^2+y^2}{2(x+y)}$$

Differentiating and simplifying yields:

$$(x^2-y^2-2xy)dx=(x^2-y^2+2xy)dy$$

Therefore, the correct option is C.

Let the center be $(h,h)$. Then the circle is

$$(x-h)^2+(y-h)^2=r^2$$

Because it passes through the origin,

$$(0-h)^2+(0-h)^2=r^2$$ $$2h^2=r^2$$

Substituting back,

$$(x-h)^2+(y-h)^2=2h^2$$

Now expand:

$$x^2-2hx+h^2+y^2-2hy+h^2=2h^2$$ $$x^2+y^2-2h(x+y)=0$$

Differentiate this relation with respect to $x$ treating $h$ as the parameter to be eliminated, and on simplification we obtain

$$(x^2-y^2-2xy)dx=(x^2-y^2+2xy)dy$$

Thus the required differential equation is the one in option C.