Question #d6ef5

Feb 11, 2017

The differential equation for the family of circles is:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{r + x}{r - y}$ where $r > 0$.

Explanation:

The general equation of a circle with centre $\left(a , b\right)$ and radius $r$ is:

${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$

If we want the circle in the second quadrant then we require the centre $\left(a , b\right)$ to lie on the line $y = - x$ so that $a = - b$ and $r = b$ with $r > 0$, which gives us:

${\left(x + r\right)}^{2} + {\left(y - r\right)}^{2} = {r}^{2} \setminus \setminus \setminus \setminus \setminus \setminus \setminus$ where $r > 0$

Differentiating wrt $x$ we get:

$2 \left(x + r\right) + 2 \left(y - r\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 0$
$\therefore \left(x + r\right) + \left(y - r\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 0$
$\therefore \left(y - r\right) \frac{\mathrm{dy}}{\mathrm{dx}} = - \left(x + r\right)$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{r + x}{r - y}$

So the differential equation for the family of circles is:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{r + x}{r - y}$ where $r > 0$.