91Ó°ÊÓ

Since the circles pass through the origin, their equation can be written in the form: \[ (x-a)^2 + y^2 = a^2 \] Here, 'a' is the x-coordinate of the center of the circle.

To find the slope at any point (x,y) on this circle, differentiate the equation with respect to \({x}\). Use implicit differentiation: \[ 2(x-a) + 2y\frac{dy}{dx} = 0 \] Now, solve for the slope \(\frac{dy}{dx}\): \[ \frac{dy}{dx} = \frac{a-x}{y} \]

For the orthogonal trajectories, the slopes of the tangent lines must be negative reciprocals of each other. Let \({m_1}\) be the slope of the tangent line to the given circle and \({m_2}\) be the slope of the tangent line to the orthogonal trajectory at the point of intersection. Then, \[ m_1 \cdot m_2 = -1 \] Since the slope of the tangent line to the given circle is \(\frac{dy}{dx} = \frac{a-x}{y}\), let the slope of the orthogonal trajectory be \(\frac{dx}{dy}\). Therefore, we have: \[ \frac{a-x}{y} \cdot \frac{dx}{dy} = -1 \]

We need to eliminate the parameter 'a' from the differential equation to get a relation between \({x}\), \({y}\), and their derivatives. To do that, let's use the equation of the circle to express 'a' in terms of \({x}\) and \({y}\) and then substitute it back into the differential equation: \[ (x-a)^2 + y^2 = a^2 \\ a^2 - 2ax +x^2 + y^2 = a^2 \\ a = \frac{x^2+y^2}{2x} \] Now substitute this expression for 'a' into the differential equation: \[ \frac{\frac{x^2+y^2}{2x} - x}{y} \cdot \frac{dx}{dy} = -1 \]

To find the equation of the orthogonal trajectories, solve the above differential equation. Here's a simplified version: \[ \frac{x^2+y^2 - 2x^2}{2xy} \cdot \frac{dx}{dy} = -1 \\ \frac{-x^2+y^2}{2xy} \cdot \frac{dx}{dy} = -1 \\ \] Separating the variables and integrate: \[ \frac{dx}{dy} = \frac{-2xy}{-x^2 + y^2} \\ \int \frac{-2xy}{-x^2 + y^2} dy = \int dx \\ \] Integrating both sides will yield: \[ \int -\frac{2xy}{y^2 - x^2} dy = \int dx \\ \] Use the substitution method by letting \(v = y^2 - x^2\) so that \(dv = 2y \cdot dy\). Then, the equation becomes: \[ \int -\frac{1}{v} dv = \int dx \\ -\ln |v| + C_1 = x + C_2 \\ \] Now substitute \(v=y^2-x^2\) back: \[ -\ln |y^2 - x^2| = x + C \\ \] Hence, the orthogonal trajectories are given by the equation: \[ -\ln |y^2 - x^2| = x + C \]