Two questions for coordinate geometry Note: I am burning through dozens of questions from sample papers and these i couldnt understand, these are not homework and i would appreciate it if the full answer could be provided. 
The first question i dont understand at all, how do you even proceed with that 0.0
Find the coordinates of the centre of circle passing through the point (0,0), (-2,1) and (-3,2). Also find its radius.
Second one also i have no idea how to solve this type of a question, the mid points of the sides of $\triangle PQR$ are (1,2), (0,1), (1,0) then find the coordinates of the three vertices of the triangle. 
 A: 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. 
A: First Question
A circle is defined by the equation $(x-x_0)^2+(y-y_0)^2 = r^2$, where $(x_0,y_0)$ is the center of the circle and $r$ is its radius (this practically says "all points $(x,y)$ that are of distance $r$ from $(x_0,y_0)$").
You are given three points on the circle, and you can plug those into the equation:
$(0,0)$: $$x_0^2+y_0^2 = r^2$$
$(-2,1)$: $$(x_0+2)^2+(y_0-1)^2 = r^2$$
$$x_0^2 +4x_0 + 4 + y_0^2 - 2y_0 + 1 = r^2$$
$$r^2 +4x_0 + 5 - 2y_0 = r^2$$
$$4x_0 - 2y_0 = -5$$
$(-3,2)$: $$(x_0+3)^2+(y_0-2)^2 = r^2$$
$$x_0^2 + 6x_0 + 9 + y_0^2 - 4y_0 + 4 = r^2$$
$$6x_0 - 4y_0 = -13$$
You can subtract twice the $(-2,1)$ equation to the $(-3,2)$ equation to get
$$x_0=1.5$$
and then substitute to get $$y_0=5.5$$
Now from the $(0,0)$ equation you get $$r=\sqrt{\frac{9}{4}+\frac{121}{4}}=\frac{1}{2}\sqrt{30}$$
I might get to the second later if I can procrastinate some more.
A: Hint: the circle whit center in $(a,b)$ and ratio $r>0$ is the set 
$$
C=\{ (x,y)\in\mathbb{R}^2 : (x-a)^2 +(y-b)^2=r^2 \}.
$$
Build a system of three quadratic equations in the variables 'a', 'b' and 'r' from the three pertinence: $(0,0)\in C$, $(-2,1)\in C$ and $(-3,2)\in C$.
To second question use de formula: 

