Finding an equation for a circle given its center and a point through which it passes No idea how to do this, I used to have these conic shapes committed to memory but I forget them already. 

I am supposed to find an equation for the circle that has center $(-1, 4)$ and passes through $(3, -2)$.

I tried graphing it, but it didn't help since the point was not in a straight line with the center and I have no idea what the diameter is.
Recall the:

 Distance Formula:

The distance between the points $(x_1,y_1)$ and $(x_2,y_2)$ is
$$
D=\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}
$$



Example: 

The radius of your circle is the distance between the points  $(-1,4)$ and $(3,-2)$. Using the Distance Formula:
$$
D= \sqrt{  \bigl(3-(-1)\bigr)^2 + (-2-4)^2  } 
= \sqrt{  4^2 + (-6)^2  } 
= \sqrt{ 16+36 }
=  \sqrt{ 52 } .
$$



What is the equation of the circle?

It is important to realize   the "equation of the circle" is: a point $(x,y)$ is on the circle if and only if the coordinates of the point $x$ and $y$ satisfy
the equation. So, how to get the equation? What is the relationship between the $x$ and $y$ coordinates of a point on the circle?

Well, let $(x,y)$ be a point on the circle. The big idea is:

$$
\text{The distance from the point }(x,y)\text{  to the center }(-1,4)\text{ is }\sqrt{52}.$$


So, using the distance formula (with $(x_2,y_2)=(x,y)$ and $(x_1,y_1)=(-1,4)$) , it follows that 
$$
\sqrt{52}=\sqrt{\bigl(x-(-1)\bigr)^2 +(y-4)^2}.
$$  

Or
$$
52=(x+1)^2 +(y-4)^2.
$$  

The shortcut would be to just use the following formula (But it's important to realize why you'd use it and where it comes from):

Equation of a Circle

The equation of the circle with center located at $(a,b)$ and with radius $r$ is
$$
r^2=(x-a)^2 +(y-b)^2
$$
Note that the radius squared is on the left-hand side of the equation. 




 The following may help:
1 find equation of the circle passing through (1,0) and (0,1) and having the smallest possible radius .
2  show that cot 15/2 degree = √2+√3+√4+√6 .