Question

Find Orthogonal Trajectories of
circles x²+ (y-c)² = c²​

Answer 1

Answer:

x²+(y+2c)y=0

Step-by-step explanation:

x²+ (y-c)² = c²

x²+y²-2yc+c²=c²

x²+(y+2c)y=0

Answer 2

Answer:

Orthogonal Trajectories of circles x²+ (y-c)² = c² is$y(x) = c_1 \times x + \frac{c_2}{x} + c$

Step-by-step explanation:

Given, equation of circles x²+ (y-c)² = c²

To find the orthogonal trajectories of the circles given by the equation x² + (y-c)² = c², we can use the following steps:

Differentiate both sides of the given equation with respect to x,

$2x + 2(y-c) \frac{dy}{dx} = 0$

Solve for dy/dx:

$\frac{dy}{dx} = -x/(y-c)$

The orthogonal trajectories will have slopes that are negative reciprocals of the slopes of the circles, so we need to invert the expression for dy/dx and change the sign:

$\frac{dy}{dx} = \frac{ (y-c)}{x}$

Differentiate the last expression with respect to x to find the differential equation for the orthogonal trajectories,

$\frac{d}{dx} \frac{ - dy}{dx}= \frac{d}{dx}[ \frac{(y-c)}{x}]</p><p> \frac{d²y}{dx²} = \frac{(y-c)}{x²}$

This is the differential equation for the orthogonal trajectories of the circles x² + (y-c)² = c². We can solve this differential equation to find the equation of the orthogonal trajectories.

One method to solve this differential equation is to use the substitution u = y-c. Then, we have du/dx = dy/dx, and the differential equation becomesd²u/dx² = u/x²

This is a linear second-order differential equation, which can be solved using standard methods. One solution is $u(x) = c_1 \times x + \frac{c_2}{x}$

where $c_1 \: and \: c_2$ are constants determined by the initial conditions. Substituting back u = y-c, we get:

$y(x) = c_1 \times x + \frac{c_2}{x} + c$

where c is another constant of integration.

Therefore, the equation of the orthogonal trajectories of the circles x² + (y-c)² = c² is given by:$y(x) = c_1 \times x + \frac{c_2}{x} + c$

Circle related two more questions:

https://brainly.in/question/53451096

https://brainly.in/question/2727823

#SPJ3