Question

Find the orthogonal trajectories of the given family of curves : all circles through the points \((1,1)\) and \((-1,-1)\).

Short answer

The equation for the orthogonal trajectories of the given family of circles is $\bar{y} = 2x + C$.

Step-by-step solution

Find the differential equation for family of circles passing through \((1,1)\) and \((-1,-1)\)

Let's consider a circle with center \((h, k)\) and radius \(r\). The equation of this circle can be written as: \((x-h)^2 + (y-k)^2 = r^2\) Since this circle passes through the points \((1,1)\) and \((-1,-1)\), we can substitute these points into the equation and obtain: \((1-h)^2 + (1-k)^2 = r^2\) and \((-1-h)^2 + (-1-k)^2 = r^2\) We can eliminate \(r^2\) by equating these two expressions: \((1-h)^2 + (1-k)^2 = (-1-h)^2 + (-1-k)^2\) After simplifying the equation above, we have: \(4(h+k) = 4\) Now we can represent the equation of the circle as: \((x-h)^2 + (y-k)^2 = r^2\) and \(h+k=1\)

Eliminate \(h\) and \(k\) to find the differential equation

Take implicit derivatives with respect to \(x\) for both equations: For the circle equation: \(2(x-h) + 2(y-k)\frac{dy}{dx} = 0\) For the second equation: \(\frac{d(h+k)}{dx} = 0 \implies \frac{dh}{dx} + \frac{dk}{dx} = 0\) Now let's eliminate h and k from the circle equation using the previous equation: \(2(x-\frac{1}{2}(2k-y)) + 2(y-\frac{1}{2}(x+1))\frac{dy}{dx} = 0\) Simplify the expression: \(2(x-y)+x\frac{dy}{dx} + y\frac{dy}{dx} = -2\) Now we'll isolate the derivative and obtain the differential equation for the family of curves: \(\frac{dy}{dx}=\frac{2(x-y)}{y-x}\)

Change the sign of the derivative to find the orthogonal trajectories

To find the orthogonal trajectories, we need to change the sign of the derivative in the differential equation: \(\frac{d\bar{y}}{dx}=-\frac{2(x-y)}{y-x}\) We can simplify it more : \(\frac{d\bar{y}}{dx}=\frac{2(y-x)}{y-x} = 2\) Now, we need to solve this differential equation for \(\bar{y}\).

Solve the differential equation for the orthogonal trajectories

We have the first-order differential equation: \(\frac{d\bar{y}}{dx}=2\) Integrate with respect to x to find \(\bar{y}\): \(\int \frac{d\bar{y}}{dx} dx = \int 2 dx\) \(\bar{y} = 2x + C\) This is the equation for the orthogonal trajectories of the given family of circles through the points \((1,1)\) and \((-1,-1)\). Each curve of this form will be orthogonal to all circles from the original family.