The differential equation corresponding to al | vedclass

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(A) Let the radius of the circle be $a$. Then the center of the circle is $(a, a)$. Hence, the equation of the circle is:$(x-a)^2 + (y-a)^2 = a^2$$x^2

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$(x-y)^2\left[1+\left(\frac{d y}{d x}\right)^2\right]=\left(x+y \frac{d y}{d x}\right)^2$

Vedclass · Answer

(A) Let the radius of the circle be $a$. Then the center of the circle is $(a, a)$. Hence, the equation of the circle is:$(x-a)^2 + (y-a)^2 = a^2$$x^2 - 2ax + a^2 + y^2 - 2ay + a^2 = a^2$$x^2 + y^2 - 2ax - 2ay + a^2 = 0$  --- $(i)$Differentiating equation $(i)$ with respect to $x$, we get:$2x + 2y \frac{dy}{dx} - 2a - 2a \frac{dy}{dx} = 0$$x + y \frac{dy}{dx} - a(1 + \frac{dy}{dx}) = 0$$a = \frac{x + y \frac{dy}{dx}}{1 + \frac{dy}{dx}}$Substituting the value of $a$ into equation $(i)$:$(x-a)^2 + (y-a)^2 = a^2$$(x-a)^2 + (y-a)^2 = (\frac{x + y \frac{dy}{dx}}{1 + \frac{dy}{dx}})^2$Note that $(x-a) = x - \frac{x + y \frac{dy}{dx}}{1 + \frac{dy}{dx}} = \frac{x + x \frac{dy}{dx} - x - y \frac{dy}{dx}}{1 + \frac{dy}{dx}} = \frac{(x-y) \frac{dy}{dx}}{1 + \frac{dy}{dx}}$And $(y-a) = y - \frac{x + y \frac{dy}{dx}}{1 + \frac{dy}{dx}} = \frac{y + y \frac{dy}{dx} - x - y \frac{dy}{dx}}{1 + \frac{dy}{dx}} = \frac{y-x}{1 + \frac{dy}{dx}}$Substituting these into $(x-a)^2 + (y-a)^2 = a^2$:$(\frac{(x-y) \frac{dy}{dx}}{1 + \frac{dy}{dx}})^2 + (\frac{y-x}{1 + \frac{dy}{dx}})^2 = (\frac{x + y \frac{dy}{dx}}{1 + \frac{dy}{dx}})^2$$(x-y)^2 (\frac{dy}{dx})^2 + (x-y)^2 = (x + y \frac{dy}{dx})^2$$(x-y)^2 [1 + (\frac{dy}{dx})^2] = (x + y \frac{dy}{dx})^2$

The differential equation corresponding to all the circles lying in the first quadrant and touching the coordinate axes is

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