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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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find the differential equation of all circles which pass through the origin and whose center lies on xaxis

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All circles passing through origin and having centre on $X-axis$ say at $(a,0)$ will have $ radius= a$
$\therefore $ General equation of such circles is
$(x-a)^2+y^2=a^2$..........(i)
This equation has only one arbitrary constant.$a$.
$\therefore$ This equation can be differentiated only once to remove the arbitrary constant $a$
Differentiating (i) on both the sides we get
$2(x-a)+2y.\large\frac{dy}{dx}=0$........(ii)
$\Rightarrow\:x-a=-y.\large\frac{dy}{dx}$
and
$a=x+y\large\frac{dy}{dx}$
Substituting the values of $(x-a) \:\:and\:\: a$ in (i) we get
$y^2.(\large\frac{dy}{dx})^2$$+y^2=(x+y\large\frac{dy}{dx})^2$
$\Rightarrow\:y^2.(\large\frac{dy}{dx})^2$$+y^2=x^2+y^2.(\large\frac{dy}{dx})^2$$+2xy.\large\frac{dy}{dx}$
$\Rightarrow\:y^2=x^2+2xy.\large\frac{dy}{dx}$
is the differential equation of the given circles.
answered Dec 13, 2013 by rvidyagovindarajan_1
 

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