The differential equation of the family of ci | vedclass

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Question

(A) The equation of a circle with center $(h, 0)$ on the $X$-axis and radius $r = |h|$ (since it touches the $Y$-axis) is $(x-h)^2 + y^2 = h^2$.Expand

Vedclass · Accepted Answer

$4(x+y \frac{dy}{dx})^2 x^2 = (x^2+y^2)^2$

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(A) The equation of a circle with center $(h, 0)$ on the $X$-axis and radius $r = |h|$ (since it touches the $Y$-axis) is $(x-h)^2 + y^2 = h^2$.Expanding this,we get $x^2 - 2xh + h^2 + y^2 = h^2$,which simplifies to $x^2 + y^2 = 2xh$.Let $h = b$. Then $x^2 + y^2 = 2bx$ ... $(i)$.Differentiating with respect to $x$,we get $2x + 2y \frac{dy}{dx} = 2b$,or $x + y \frac{dy}{dx} = b$ ... $(ii)$.Substituting $(ii)$ into $(i)$,we get $x^2 + y^2 = 2x(x + y \frac{dy}{dx})$.Squaring both sides,we get $(x^2 + y^2)^2 = 4x^2(x + y \frac{dy}{dx})^2$.

The differential equation of the family of circles,whose centres are on the $X$-axis and which touch the $Y$-axis,is

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