$f(z)=\frac{(iz+2)}{(4z+i)}$ maps the real axis in the $\mathbb{C}$-plane into a circle Find the center and radius of the circle. Also find the points on the complex plane which is mapped onto the center of the circle.

Let $f(z)=w=x+iy$ and show that $(x,y)$ is on a circle, provided $z \in \mathbb{R}$.  First, solve for $z$ so we can get a look at its real and imaginary parts:  $$w=\frac{(iz+2)}{(4z+i)} \\
(4z+i)w= iz+2 \\
z(4w-i)=2-iw \\
z=\frac{2-iw}{4w-i} = \frac{2-iw}{4w-i}\frac{4\overline{w}+i}{4\overline{w}+i}=\frac{8\overline{w}+2i-4i|w|^2+w}{|4w-i|^2}$$
Since $z$ is a real number, the imaginary part of the numerator of that last expression must be zero.  So
$$-8y+2-4(x^2+y^2)+y=0 \\
x^2 + y^2 +\frac{7}{4}y=\frac{1}{2} \\
x^2 + y^2 +\frac{7}{4}y + \left(\frac{7}{8}\right)^2=\frac{1}{2}+ \left(\frac{7}{8}\right)^2 = \frac{81}{64} \\
x^2 +\left(y+ \frac{7}{8}\right)^2=\left(\frac{9}{8}\right)^2 $$
Now claim the point $f(z)=(x,y)$ does lie on a circle, the circle has center $(0,-7/8)$ and the radius of the circle is $\frac{9}{8}$.
Verified by plotting some points.
To find the point $z_c$ that $f$ maps to the center of the circle $w_c$, we can use our simplest expression for $z$ in terms of $w$ from above. 
$$z_c=\frac{2-iw_c}{4w_c-i}=\frac{2-i\left( \frac{-7}{8}i \right)}{4\left( \frac{-7}{8}i \right)-i}=\frac{2-\frac{7}{8}}{-\frac{28}{8}i-i}=\frac{i}{4}$$
