widdonod1t

2021-12-27

Differential Equations (Families of Curves): Circles with fixed radius r and tangent to the y-axis.

poleglit3

Expert

The general equation of a circle with centre at (h, k) and radius r is given by
${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={r}^{2}$
Also, it is given that y axis is tangent to this circle.
This means the perpendicular distance from the centre of the circle to the line $x=0$ is equal to the radius of the circle.
This means $h=r$
Hence, the equation of such a circle becomes
${\left(x-r\right)}^{2}+{\left(y-k\right)}^{2}={r}^{2}$
Here, k is hte only variable left.
Hence, to obtain the differential equation, we need to eliminate the variable k
We will do this with the help of differentiation.
Differentiating the curve with respect to x, we get
$2\left(x-r\right)+2\left(y-k\right)\frac{dy}{dx}=0$
$\left(y-k\right)\frac{dy}{dx}=r-x$
$\left(y-k\right)=\frac{r-x}{\frac{dy}{dx}}$
Putting the value of $y-k$ back in the equation we get
${\left(x-r\right)}^{2}+{\left(y-k\right)}^{2}={r}^{2}$
${\left(x-r\right)}^{2}+\left(\frac{r-x}{\frac{dy}{dx}}\right)={r}^{2}$
${\left(\frac{dy}{dx}\right)}^{2}{\left(x-r\right)}^{2}+{\left(x-r\right)}^{2}={\left(\frac{dy}{dx}\right)}^{2}\left({r}^{2}\right)$
${\left(\frac{dy}{dx}\right)}^{2}\left({x}^{2}+{r}^{2}-2xr-{r}^{2}\right)+{\left(x-r\right)}^{2}=0$
${\left(\frac{dy}{dx}\right)}^{2}\left({x}^{2}-2xr\right)+{\left(x-r\right)}^{2}=0$
${\left(\frac{dy}{dx}\right)}^{2}=\frac{{\left(x-r\right)}^{2}}{2xr-{x}^{2}}$

Marcus Herman

Expert

$⇒{\left(x-a\right)}^{2}+{y}^{2}={r}^{2}$
$⇒2\left(x-a\right)+2y{y}_{1}=0$
$⇒\left(x-a\right)=-y{y}_{1}$
$⇒{y}^{2}{y}_{1}^{2}+{y}^{2}={r}^{2}$

Vasquez

Expert

Center is (0,k) and radius is r.
Equation is
${x}^{2}+\left(y-k{\right)}^{2}={r}^{2}$ Equation 1
By differentiating we get,
$2x+2\left(y-k\right)\frac{dy}{dx}=0$
$k=+x\frac{dx}{dy}+y$ Equation 2
Putting the value of K from Equation 1 and 2, we get
$\begin{array}{}{x}^{2}+\left(y-x\frac{dx}{dy}-y{\right)}^{2}={r}^{2}\\ {x}^{2}+{x}^{2}\left(\frac{dx}{dy}{\right)}^{2}={r}^{2}\\ 1+\left(\frac{dx}{dy}{\right)}^{2}=\frac{{r}^{2}}{{x}^{2}}\\ \left(\frac{dy}{dx}{\right)}^{2}=\frac{{x}^{2}}{{r}^{2}-{x}^{2}}\end{array}$

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