Find the differential equation of: A family of circles passing through the origin and having centers on the $x-$axis.
I can't figure out the equation of "family of circles passing through the origin and having centers on $x-$axis",please help me..
[Math] Differential equations of family of circles
geometryordinary differential equations
Related Solutions
For a circle polar coordinates is a natural choice. It avoids differential relations between $ (x,\theta) $. It is convenient to find a pure polar equation of a circle of arbitrary radius (or diameter D ) whose inclination $ \alpha$ to a reference direction like x-axis is arbitrary. Curvature is a second order differential and hence these arbitrary constants $ ( D, \alpha) $ need to be involved and later on eliminated in differentiation:
From the sketch it is seen they are relatively positioned and geometrically related as:
$$ r = D \cos(\theta + \alpha) $$
whose differential equation is the well known:
$$ \boxed {\frac{d^2 r}{d \theta^{2}} + r =0.} $$
Rectangular coordinates to be computed after numerical integration.
EDIT 1:
In rectangular coordinates we can derive as follows. $\alpha$ can be taken wlog as zero as seen above, anyhow it disappears.
$$ x = D \cos ^2\theta=\frac{D\,x^2}{x^2+y^2} $$
$$ x^2 + y^2 = D\,x $$
Differentiating once
$$ x + y\,y^{\prime} = D/2 $$ Differentiate again
$$\boxed {y\, y^{\prime \,\prime} + 1 + y^{\prime\, 2 }=0.}$$
One disadvantage with rectangular coordinates numerical integration is that it cannot proceed beyond point of the vertical tangent.. unless other tricks are used.
The circle's centre will be $(k, 5)$ So the equation is $$(x-k)^2 + (y-5)^2 = 25.$$
Now put $(0,8)$ in the equation, to get $k = \pm 4$.
Best Answer
When the circle passes through the origin and centre lies on $x-$axis, then the abscissa will be equal to the radius of the circle and the $y-$co-ordinate of the centre will be zero. i.e., $h = a$ and $k = 0$.
Then the equation of the circle satisfied the given condition is$$(x - a)^2 + y^2 = a^2\tag1$$
Differentiating both side of $(1)$ with respect to $x$ , $$2(x-a)+2yy'=0\implies x-a=-yy'\implies a=x+yy'$$where $~y'\equiv\dfrac{dy}{dx}~$.
Putting the value of $~a~$, in equation $(1)$ , $$y^2~y'^2 + y^2 = (x+yy')^2$$ $$\implies y^2~y'^2 + y^2 = x^2+2xyy'+y^2y'^2$$ $$\implies y^2 = x^2+2xyy'$$ which is the required differential equation of the given family of circles.