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

Question

The differential equation of all circles passing through the origin and having their centres on the X-axis is (A) x2=y2+x y (d y/d x) (B) x2=y2+3 x y (d y/d x) (C) y2=x2+2 x y (d y/d x) (D) y2=x2-2 x y (d y/d x)

Tardigrade Admin · Accepted Answer

The equation of all circles through the origin and having their centres on the

X

-axis is

(x - g)^{2} + (y - 0)^{2} = g^{2}

\Rightarrow x^{2} + y^{2} - 2 gx = 0.....

(i)
On differentiating equation (i) w.r.t.

x

, we get

2 x + 2 y \frac{d y}{d x} - 2 g = 0

\Rightarrow 2 g = 2 (x + y \frac{d y}{d x})

On putting the value of

2 g

in Eq. (i), we get

x^{2} + y^{2} - 2 (x + y \frac{d y}{d x}) x = 0

\Rightarrow y^{2} - x^{2} - 2 x y \frac{d y}{d x} = 0

\Rightarrow y^{2} = x^{2} + 2 x y \frac{d y}{d x}

Q. The differential equation of all circles passing through the origin and having their centres on the $X$ -axis is

$x^{2} = y^{2} + x y \frac{d y}{d x}$

$x^{2} = y^{2} + 3 x y \frac{d y}{d x}$

$y^{2} = x^{2} + 2 x y \frac{d y}{d x}$

$y^{2} = x^{2} - 2 x y \frac{d y}{d x}$

Q. The differential equation of all circles passing through the origin and having their centres on the X-axis is

x2=y2+xydxdy​

x2=y2+3xydxdy​

y2=x2+2xydxdy​

y2=x2−2xydxdy​

Q. The differential equation of all circles passing through the origin and having their centres on the $X$ -axis is

$x^{2} = y^{2} + x y \frac{d y}{d x}$

$x^{2} = y^{2} + 3 x y \frac{d y}{d x}$

$y^{2} = x^{2} + 2 x y \frac{d y}{d x}$

$y^{2} = x^{2} - 2 x y \frac{d y}{d x}$