socorraob

2021-12-28

Family of Curves
Try and obtain the differential equations of the given families of curves
All circles passing through the origin and (0,4).

puhnut1m

We have to find the differential equation of family of curve.
Equation of circle is:
${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={r}^{2}$ (1)
9t passes through (0,0)
So, eq (1) becomes
${h}^{2}+{k}^{2}={r}^{2}$
$⇒{\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={h}^{2}+{k}^{2}$ (2)
Equation (1) also pass through (0,4)
So, ${h}^{2}+{\left(4-k\right)}^{2}={r}^{2}$
$\mathrm{¬}\left\{{h}^{2}\right\}+{\left(4-k\right)}^{2}=\mathrm{¬}\left\{{h}^{2}\right\}+{k}^{2}$
${\left(4-k\right)}^{2}={k}^{2}$
$4-k=k$
$4=2k$
$k=2$
So, equation (2) becomes
${\left(x-h\right)}^{2}+{\left(y-2\right)}^{2}={h}^{2}+4$
${x}^{2}+\mathrm{¬}\left\{{h}^{2}\right\}-2xh+{y}^{2}+\mathrm{¬}\left\{4\right\}-4y=\mathrm{¬}\left\{{h}^{2}\right\}+\mathrm{¬}\left\{4\right\}$
${x}^{2}-2xh+{y}^{2}-4y=0$ (3)
Differentiating w.r.t. x,
$2x-2h+2y\frac{dy}{dx}-4\frac{dy}{dx}=0$
$h=x+\frac{dy}{dx}\left(y-2\right)$ (4)
Put in (3),
${x}^{2}-2{x}^{2}-2xy\frac{dy}{dx}+4x\frac{dy}{dx}+{y}^{2}-4y=0$
${y}^{\prime }\left(4x-2xy\right)-{x}^{2}+{y}^{2}-4y=0$

Maricela Alarcon

All circles passing through the origin and (0, 4).
Let the equation of the circle be ${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={r}^{2}$.
Now the circle passes through the origin, therefore we get
${\left(0-h\right)}^{2}+{\left(0-k\right)}^{2}={r}^{2}$
$⇒{h}^{2}+{k}^{2}={r}^{2}\to \left(1\right)$
And also the circle passes through the point (0, 4), therefore we get
${\left(0-h\right)}^{2}+{\left(4-k\right)}^{2}={r}^{2}$
$⇒{h}^{2}+16-8k+{k}^{2}={r}^{2}$
$⇒{h}^{2}+{k}^{2}+16-8k={r}^{2}$
$⇒{r}^{2}+16-8k={r}^{2}$ [By using equation (1)]
$⇒16-8k=0$
$⇒8k=16$
$⇒k=2$
Now put $k=2$ in equation (1), we get
${h}^{2}+{2}^{2}={r}^{2}$
$RIghtarrow{r}^{2}={h}^{2}+4$
Therefore equation of the circle becomes,
${\left(x-h\right)}^{2}+{\left(y-2\right)}^{2}={h}^{2}+4$
$⇒{x}^{2}-2hx+{h}^{2}+{y}^{2}-4y+4={h}^{2}+4$
$⇒{x}^{2}-2hx+{y}^{2}-4y=0$
Now differentiating with respect to x, we get
$2x-2h\left(1\right)+2y\cdot \frac{dy}{dx}-4\cdot \frac{dy}{dx}=0$
$⇒\frac{dy}{dx}\left(2y-4\right)=2h-2x$
$⇒\frac{dy}{dx}=\frac{2h-2x}{2y-4}$
$⇒\frac{dy}{dx}=\frac{\frac{{x}^{2}+{y}^{2}-4y}{x}-2x}{2y-4}$

karton