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

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Solution

Q.
The differential equation of the family of circles passing through the origin and having centre at the line y = x is [2024]

1 $(x^{2} + y^{2} + 2 x y) d x = (x^{2} + y^{2} - 2 x y) d y$

2 $(x^{2} - y^{2} + 2 x y) d x = (x^{2} - y^{2} - 2 x y) d y$

3 $(x^{2} + y^{2} - 2 x y) d x = (x^{2} + y^{2} + 2 x y) d y$

4 $(x^{2} - y^{2} + 2 x y) d x = (x^{2} - y^{2} + 2 x y) d y$

Ans.
(2)

Let (k, k) be the centre of circle, so equation of circle with radius r is given by

${(x - k)}^{2} + {(y - k)}^{2} = r^{2}$

Now, circle is passing through origin

$\Rightarrow r^{2} = 2 k^{2}$

So, equation of circle becomes

$x^{2} + y^{2} - 2 k x - 2 k y = 0$ ... (i)

On differentiating (i) w.r.t. x, we get

2x + 2yy' – 2k – 2ky' = 0

$\Rightarrow k = \frac{x + y y'}{(1 + y')}$

Substituting the value of k in (i) we get

$x^{2} + y^{2} - 2 x (\frac{x + y y'}{1 + y'}) - 2 y (\frac{x + y y'}{1 + y'}) = 0$

$\Rightarrow (x^{2} + y^{2}) (1 + y') - 2 x^{2} - 2 x y y' - 2 x y - 2 y^{2} y' = 0$

$\Rightarrow - x^{2} + y^{2} + x^{2} y' - y^{2} y' - 2 x y y' - 2 x y = 0$

$\Rightarrow (x^{2} - y^{2} - 2 x y) y' = x^{2} - y^{2} + 2 x y$

$\Rightarrow (x^{2} - y^{2} - 2 x y) d y = (x^{2} - y^{2} + 2 x y) d x$

is the required differential equation.

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