Answer to Question #263691 in Differential Equations for Lotus

Question

Obtain the differential equation that describe the family of curve.

1. All straight lines tangent to a unit circle with center at (1, 1)

Answer 1

y = mx + c will be a tangent to a unit circle with center (0,0) if perpendicular distance of it from origin is 1.

So $\frac{|0-m.0-c|}{\sqrt{1²+m²}}=1$

=> c = ±$\sqrt{1+m²}$

So equation of tangent is

y = mx ± $\sqrt{1+m²}$

Here circle has center at (1,1) and radius 1 unit

Let us translate the origin to (1,1) and the reduced equation of the circle will be X² + Y² = 1 where X = x-1 and Y = y-1

So the equation of tangent in new coordinate system will be

Y = mX ± $\sqrt{1+m²}$ , m is a arbitrary constant.

Differentiating with respect to x

$\frac{dY}{dX} = m$

Eleminating the arbitrary constant m we get the differential equation of tangent as

Y = $\frac{dY}{dX}X±\sqrt{1+(\frac{dY}{dX})²}$

Since Y = y-1 and X = x-1 , $\frac{dY}{dX} = \frac{dy}{dx}$

So the differential equation of tangent to the unit circle with center (1,1) is

y - 1 = (x-1)$\frac{dy}{dx}±\sqrt{1+(\frac{dy}{dx})²}$

=> [ y - 1 - (x-1)$\frac{dy}{dx} ]²$=$[\sqrt{1+(\frac{dy}{dx})²}$ ]²

^=> [ y - 1 - (x-1)$\frac{dy}{dx} ]² = 1 + (\frac{dy}{dx})²$