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.
 A: The equation of a circle with center $(a,b)$ and radius $r$ is
$$(x-a)^2 + (y-b)^2 = r^2.$$
You know the center. You also know one point on the circle. The radius is the distance from the center of the circle to any one point on the circle.
So find the distance between $(-1,4)$ and $(3,-2)$ to get the radius. Then use the radius and the center to get the equation. 
The diameter is, of course, twice the radius, but is irrelevant here.
A: In general a circle with center $(h,k)$ and radius $r$ can be expressed as:
$$(x-h)^2+(y-k)^2=r^2$$
In the case of our questions, you have $(h,k)=(-1,4)$ in the above equation. You can then substitute $(-3,2)$ for $(x,y)$ to find $r$.
A: 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: 
 
A: Well you should know (or look up in your textbook) that a circle is described by the equation  $(x - x_0)^2 + (y - y_0)^2 = r^2$ where $(x_0, y_0)$ is the center and $r$ is the radius.  So you see your circle should be $(x +1)^2 + (y - 4)^2 = r^2$.  The only thing missing is $r^2$.  But the problem tells you (3, -2) is a point on the circle.  So plugging this into your equation, you will find your answer.
A: The equation of a circle with center $(a,b)$ and radius $r$ is
$$(x-a)^2 + (y-b)^2 = r^2.$$
Because the center is $(a,b)=(-1,4)$ so $a=-1,b=4$. You also know one point on the circle. The radius is the distance from the center of the circle to any one point on the circle.
The distance between $(-1,4)$ and $(3,-2)$ is $r=\sqrt52$ or $r^2=52$. Then use the radius and the center to get the equation. 
$$(x+1)^2 + (y-4)^2 = 52$$
A: The smallest circle passing through those points  would be the circle with a diameter from $(1,0)$ to $(0,1)$.    It therefore has a radius of $\frac{\sqrt2}{2}$ with a center of $(\frac12,\frac12)$.   So,     $(x - \frac{1}{2})^2 + (y - \frac12)^2 = \frac12$  would satisfy all conditions.    
Any radius smaller than that could not include the two poinst and anything larger than that would not fit the problem.  
A: 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 .
