Convert the circle equation 4x2 + 4y2 − 6x + 10y − 1 = 0 to standard form and find its center and radius.

The given equation of the circle is $4x^2 + 4y^2 - 6x + 10y - 1 = 0$.
To convert this to the standard form $(x-h)^2 + (y-k)^2 = r^2$, where $(h,k)$ is the center and $r$ is the radius, first divide the entire equation by 4:

$$x^2 + y^2 - \frac{6}{4}x + \frac{10}{4}y - \frac{1}{4} = 0$$ $$x^2 + y^2 - \frac{3}{2}x + \frac{5}{2}y - \frac{1}{4} = 0$$

To complete the square for $x^2 - \frac{3}{2}x$, add $(\frac{1}{2} \cdot -\frac{3}{2})^2 = (-\frac{3}{4})^2 = \frac{9}{16}$.
To complete the square for $y^2 + \frac{5}{2}y$, add $(\frac{1}{2} \cdot \frac{5}{2})^2 = (\frac{5}{4})^2 = \frac{25}{16}$.
Add these values to both sides of the equation:

$$(x^2 - \frac{3}{2}x + \frac{9}{16}) + (y^2 + \frac{5}{2}y + \frac{25}{16}) = \frac{1}{4} + \frac{9}{16} + \frac{25}{16}$$

$$(x - \frac{3}{4})^2 + (y + \frac{5}{4})^2 = \frac{4}{16} + \frac{9}{16} + \frac{25}{16}$$ $$(x - \frac{3}{4})^2 + (y + \frac{5}{4})^2 = \frac{4+9+25}{16}$$ $$(x - \frac{3}{4})^2 + (y + \frac{5}{4})^2 = \frac{38}{16}$$ $$(x - \frac{3}{4})^2 + (y + \frac{5}{4})^2 = \frac{19}{8}$$

Comparing $(x - \frac{3}{4})^2 + (y + \frac{5}{4})^2 = \frac{19}{8}$ with $(x-h)^2 + (y-k)^2 = r^2$:
The center $(h,k)$ is $(\frac{3}{4}, -\frac{5}{4})$.
The radius squared $r^2$ is $\frac{19}{8}$.
The radius $r$ is $\sqrt{\frac{19}{8}}$.
To simplify the radius, rationalize the denominator:

$$r = \sqrt{\frac{19}{8}} = \frac{\sqrt{19}}{\sqrt{8}} = \frac{\sqrt{19}}{2\sqrt{2}} = \frac{\sqrt{19} \cdot \sqrt{2}}{2\sqrt{2} \cdot \sqrt{2}} = \frac{\sqrt{38}}{2 \cdot 2} = \frac{\sqrt{38}}{4}$$

The center of the circle is $\boxed{\left(\frac{3}{4}, -\frac{5}{4}\right)}$ and the length of the radius is $\boxed{\frac{\sqrt{38}}{4}}$.