$\newcommand{\dd}{\partial}\newcommand{\Reals}{\mathbf{R}}\newcommand{\Basis}{\mathbf{e}}\newcommand{\Sph}{\mathbf{r}}$A mathematician might denote spherical coordinates by
$$
\left[\begin{array}{@{}c@{}}
x \\
y \\
z \\
\end{array}\right]
= \Sph(\rho, \theta, \phi)
= \left[\begin{array}{@{}c@{}}
\rho\cos\theta\sin\phi \\
\rho\sin\theta\sin\phi \\
\rho\cos\phi \\
\end{array}\right].
$$
The derivative $D\Sph$ is represented by the matrix of partial derivatives
$$
\left[\begin{array}{@{}ccc@{}}
\frac{\dd x}{\dd \rho} & \frac{\dd x}{\dd \theta} & \frac{\dd x}{\dd \phi} \\
\frac{\dd y}{\dd \rho} & \frac{\dd y}{\dd \theta} & \frac{\dd y}{\dd \phi} \\
\frac{\dd z}{\dd \rho} & \frac{\dd z}{\dd \theta} & \frac{\dd z}{\dd \phi} \\
\end{array}\right]
= \left[\begin{array}{@{}c@{}}
\cos\theta\sin\phi & -\rho\sin\theta\sin\phi & \rho\cos\theta\cos\phi \\
\sin\theta\sin\phi & \rho\cos\theta\sin\phi & \rho\sin\theta\cos\phi \\
\cos\phi & 0 & -\rho\sin\phi \\
\end{array}\right].
$$
The (normalized) columns
$$
\frac{\dd\Sph}{\dd \rho} = D\Sph\, \Basis_{\rho},\qquad
\frac{1}{\rho}\,\frac{\dd\Sph}{\dd \theta} = \frac{1}{\rho}\,D\Sph\, \Basis_{\theta},\qquad
\frac{1}{\rho}\,\frac{\dd\Sph}{\dd \phi} = \frac{1}{\rho}\,D\Sph\, \Basis_{\phi}
$$
are the "coordinate vector fields" for spherical coordinates.
In the diagram, the red and blue curves lie in a coordinate surface $\rho = \rho_{0}$ (i.e., a sphere); the blue and green curves lie in a coordinate surface $\theta = \theta_{0}$ (a longitudinal plane); the red and green curves lie in a coordinate surface $\phi = \phi_{0}$ (a cone about the $z$-axis). Small portions of the coordinates curves for $\rho$, $\theta$, and $\phi$ are green, red, and blue respectively. The coordinate fields (not shown) are unit tangent fields along the respective curves.
Generally, if $F$ is an arbitrary change of coordinates (in space, say), then the domain of $F$ is some open subset $U$ of $\Reals^{3}$, and $F$ maps $U$ "diffeomorphically" (smoothly and with smooth inverse) into $\Reals^{3}$. If we write
$$
(x, y, z) = F(u, v, w),
$$
then for each $u_{0}$ we may "restrict" $F$, obtaining a parametric surface $(v, w) \mapsto F(u_{0}, v, w)$. The image of this mapping is the "coordinate surface" $u = u_{0}$.
Similarly, we might consider $(u, w) \mapsto F(u, v_{0}, w)$ for some $v_{0}$, or $(u, v) \mapsto F(u, v, w_{0})$ for some $w_{0}$. The intersection of two coordinate surfaces, say $v = v_{0}$ and $w = w_{0}$, is the "coordinate curve" $u \mapsto F(u, v_{0}, w_{0})$. This parametric curve has a normalized velocity vector at each point, the $u$-coordinate vector field along the curve.
Incidentally, the notation suggests you're reading an engineering text. One cultural difference between engineers and mathematicians is that:
Engineers tend to denote mappings (functions) by assigning letters to input and output values, which can lead to profusions of letters (as in $a, b, c$, $m_{1}, m_{2}, m_{3}$, $r, \theta, \phi$, $R, \gamma, \beta$).
Mathematicians tend to focus on the functional relationship between inputs and outputs, which sometimes goes so far as to actively suppress the names of input and output variables (as in $(x, y, z) = \Sph(\rho, \theta, \phi)$, and speaking of $D\Sph$ instead of the partial derivatives $\dd x/\dd \rho$, etc.)
Among other things, engineering notation tends to be global: Each quantity has a single symbol, sometimes across an entire discipline.
By contrast, mathematical notation tends to be local and context-dependent: The meanings of symbols can change even from paragraph to paragraph, though symbols are chosen to obey loose cultural assumptions. (The beginning of the alphabet ($a$ through $c$) is generally reserved for constants, the middle ($i$ through $n$) for discrete (integer) indices, the late middle ($r$ through $t$ or $w$) for continuous parameters, the end ($x$ through $z$) for coordinate functions.)
Each notational convention has advantages and disadvantages. The better you're able to understand both (even if you spend most of your career in one camp or the other), the easier a time you'll have with the literature (textbooks and papers).
Best Answer
Circle Problem: There are a couple of approaches. We could use the equation of the general circle. There is a somewhat more geometric way, which makes the algebra a little simpler.
The line segment through $(0,0)$ and $(-2,1)$ has slope $-\frac{1}{2}$. The midpoint $M$ of this line segment has coordinates $(-1,1/2)$.
Imagine drawing the line through $M$ perpendicular to the line segment through $(0,0)$ and $(-2,1)$. This line passes through the centre and has slope the negative reciprocal of $-\frac{1}{2}$, which is $2$. So the line has equation $$\frac{y-1/2}{x+1}=2,$$ which simplifies to $y=2x+5/2$.
Do the same with the points $(0,0)$ and $(-3,2)$. The slope is $-\frac{2}{3}$, and the midpoint $N$ is $(-3/2,1)$. Draw the line through $N$ perpendicular to the line segment. This line has slope $\frac{3}{2}$, so it has equation $$\frac{y-1}{x+3/2}=\frac{3}{2},$$ which simplifies to $y=(3/2)x+13/4$.
The centre is the point in common between our two lines. So solve the system of $2$ linear equations. We get $x=\frac{3}{2}$ and $y=\frac{11}{2}$.
Now that we have the centre, we can easily find the radius, by computing the distance from the centre to, say, $(0,0)$. The answer turns out to be $\frac{\sqrt{130}}{2}$.
Triangle Problem: Draw a picture of a triangle, and the midpoints of its sides. Join these midpoints to each other. Note that we get a triangle whose sides are parallel to the sides of the mother triangle.
Suppose these midpoints are $(1.2)$, $(1,0)$, and $(0,1)$. For this choice of midpoints, a careful diagram would do the job, but we proceed as if the given points were less nice.
Calculate the three slopes of the lines through pairs of these points. We get slopes "$\infty$" (vertical line), $1$, and $-1$. So the three sides of the mother triangle have these slopes. That means that the three lines that form these sides have equations of shape $x=a$, $y=-x+b$, and $y=x+c$.
To find $a$, $b$, and $c$, note for example that the vertical side of the mother triangle has to go through $(0,1)$. So $a=0$. The side $y=-x+b$ has to go through $(1,2)$, so $b=3$. And the side $y=x+c$ has to go through $(1,0)$, so $c=-1$.
Now we know the equations of the sides of the mother triangle, so can find the vertices.