Technically the path is independent of the speed, what you're talking about is a parametrization of the path. And even then, your description does not determine a single parametrization because you did not specify where on the path the parametrization should start, which way around the circle we want to go, and how many times we should loop around.

Nonetheless, we can find a parametrization that fits this bill. Firstly, let's look at the "standard" parametrization of the circle of radius 5 centered at the origin going around at a speed of \omega exactly once:

\(\displaystyle{r}{\left({t}\right)}={5}{\cos{{\left(\omega{t}\right)}}}{i}+{5}{\sin{{\left(\omega{t}\right)}}}{j},{t}\in{\left[{0},\omega{2}\times\pi\right)}\)

To tilt this on its side so that it's in the y=0 plane instead of the z=0 plane, we can just replace j with k:

\(\displaystyle{r}{\left({t}\right)}={5}{\cos{{\left(\omega{t}\right)}}}{i}+{5}{\sin{{\left(\omega{t}\right)}}}{k},{t}\in{\left[{0},{2}\pi\omega\right)}\)

To have it go at a speed of 15 radians per unit of time, we just insert 15 for \omega :

\(\displaystyle{r}{\left({t}\right)}={5}{\cos{{\left({15}{t}\right)}}}{i}+{5}{\sin{{\left({15}{t}\right)}}}{k},{t}\in{\left[{0},{2}\pi{15}\right)}\)

\(\displaystyle{r}{\left({t}\right)}={5}{\cos{{\left({15}{t}\right)}}}{i}+{5}{\sin{{\left({15}{t}\right)}}}{k},{t}\in{\left[{0},{152}\pi\right)}\)

And finally to move it so that it's center is at (−7,1,1) (which incidentally will also move it from the y=0 plane to the y=1 plane), we just add the constant vector −7 i+j+k to it:

\(\displaystyle{r}{\left({t}\right)}={\left[−{7}+{5}{\cos{{\left({15}{t}\right)}}}\right]}{i}+{j}+{\left[{1}+{5}{\sin{{\left({15}{t}\right)}}}\right]}{k},{t}\in{\left[{0},{2}\pi{15}\right)}\)

\(\displaystyle{r}{\left({t}\right)}={\left[−{7}+{5}{\cos{{\left({15}{t}\right)}}}\right]}{i}+{j}+{\left[{1}+{5}{\sin{{\left({15}{t}\right)}}}\right]}{k},{t}\in{\left[{0},{152}\pi\right)}\)

And we're done!

Nonetheless, we can find a parametrization that fits this bill. Firstly, let's look at the "standard" parametrization of the circle of radius 5 centered at the origin going around at a speed of \omega exactly once:

\(\displaystyle{r}{\left({t}\right)}={5}{\cos{{\left(\omega{t}\right)}}}{i}+{5}{\sin{{\left(\omega{t}\right)}}}{j},{t}\in{\left[{0},\omega{2}\times\pi\right)}\)

To tilt this on its side so that it's in the y=0 plane instead of the z=0 plane, we can just replace j with k:

\(\displaystyle{r}{\left({t}\right)}={5}{\cos{{\left(\omega{t}\right)}}}{i}+{5}{\sin{{\left(\omega{t}\right)}}}{k},{t}\in{\left[{0},{2}\pi\omega\right)}\)

To have it go at a speed of 15 radians per unit of time, we just insert 15 for \omega :

\(\displaystyle{r}{\left({t}\right)}={5}{\cos{{\left({15}{t}\right)}}}{i}+{5}{\sin{{\left({15}{t}\right)}}}{k},{t}\in{\left[{0},{2}\pi{15}\right)}\)

\(\displaystyle{r}{\left({t}\right)}={5}{\cos{{\left({15}{t}\right)}}}{i}+{5}{\sin{{\left({15}{t}\right)}}}{k},{t}\in{\left[{0},{152}\pi\right)}\)

And finally to move it so that it's center is at (−7,1,1) (which incidentally will also move it from the y=0 plane to the y=1 plane), we just add the constant vector −7 i+j+k to it:

\(\displaystyle{r}{\left({t}\right)}={\left[−{7}+{5}{\cos{{\left({15}{t}\right)}}}\right]}{i}+{j}+{\left[{1}+{5}{\sin{{\left({15}{t}\right)}}}\right]}{k},{t}\in{\left[{0},{2}\pi{15}\right)}\)

\(\displaystyle{r}{\left({t}\right)}={\left[−{7}+{5}{\cos{{\left({15}{t}\right)}}}\right]}{i}+{j}+{\left[{1}+{5}{\sin{{\left({15}{t}\right)}}}\right]}{k},{t}\in{\left[{0},{152}\pi\right)}\)

And we're done!