Question

Find the family of orthogonal trajectories of the circles ${\left(x-h\right)}^{\mathbf{2}}\mathbf{+}{\mathit{y}}^{\mathbf{2}}\mathbf{=}{\mathit{h}}^{\mathbf{2}}$. (See the instructions above Problem 2.31.)

Short answer

Answer

The family of orthogonal trajectories of the circles:

Step-by-step solution

Given information

The given equation of the circle is${\left(x-h\right)}^2+y^2=h^2$.

Standard equation of a circle

The standard equation of a circle is ${\left(x-h\right)}^{\mathbf{2}}\mathbf{+}{\left(y-k\right)}^{\mathbf{2}}\mathbf{=}{\mathit{r}}^{\mathbf{2}}$

Here, $\left(h,k\right)$is the center of the circle and r is the radius of the circle.

The orthogonal trajectories of the circle

The given equation of the circle is,

${\left(x-h\right)}^2+y^2=h^2$

The standard equation of a circle is

${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2............\left(i\right)$

where $\left(h,k\right)$ is the center of the circle and r is the radius of the circle.

Now, compare equation (1) with the standard form of equation, which indicates that

$\left(h,0\right)$is the center of the circle and h is the radius of the circle.

Therefore, it can be said that the equation (1) shows a family of circles with center $\left(h,0\right)$and radius h. The value of h varies from $\infty$to $\infty$.