Find center, radius and a tangent to $x^2+y^2+6x-4y+3=0$ 
For the circle  $x^2+y^2+6x-4y+3=0$ find
  a) The center and radius
  b) The equation of the tangent line at the point $(-2,5)$  

Now, I solved a) and got the equation
$$(x+3)^2+(y-2)^2=10$$ with center $=(-3,2)$ and radius $=\sqrt{10}$
Now, I've never learned about the tangent of a circle, but I think that it's a line that touches the outer end of a circle. But I'm not 100% on that. So if anyone can help me out with this, that would be very beneficial. And please do not solve this for me.
 A: The line tangent to the circle at $(-2,5)$ is the straight line through $(-2,5)$ that is perpendicular to the radius of the circle that runs from the centre of the circle at $(-3,2)$ to the point $(-2,5)$. Find the slope of that radius, use that to get the slope of the tangent line, and you’re on your way.
Edit: I just corrected the $x$-coordinate of the centre. Yours has the wrong sign: $x+3=x-(-3)$, so the correct $x$-coordinate is $-3$.
A: Let us complete the square. We get $x^2+6*x+9+y^2-4*y+4-9-4+3=0$ or $(x+3)^2+(y-2)^2=10$. Your circle is with center $O(-3,2)$ and radius $r=\sqrt{10}$ 
For the second question, let us calculate slope, which would  be $\text{rise}/\text{run}$ or in our case center is $(-3,2)$  so slope=$(5-2)/(-2+3)=3$  slope of  tangent line would be $-1/3$ put this conditions into form of tangent line $y=mx+b$ we get $5=-1/3\cdot(-2)+b$  or $b=5-2/3=13/3$
We get equation of tangent line
$y=-1/3\cdot x+13/3$ 
A: Since the question has already been settled nicely, let me demonstrate an unconventional way to find the tangent of a circle through a given point on the circle.
There are three things that allow us to do what we will be doing in the sequel:


*

*Any point can be thought of as a "point-circle"; that is, a circle of radius $\textbf 0$.

*One can always produce the equation for the radical line of two circles.

*The radical line of two tangent circles is their common tangent line.
Here, now, is the procedure. We are given the point $(-2,5)$ on the circle $x^2+y^2+6x-4y+3=0$. Constructing the equation of a point-circle is a snap (note that as always, we negate the coordinates of the center when inserting them into the equation of a circle):
$$(x+2)^2+(y-5)^2=0$$
which in expanded form, looks like
$$x^2+y^2+4x-10y+29=0$$
From the equivalence of the tangent line and the radical line in this case that I mentioned earlier, all we have to do to find the tangent line to your circle is to find the radical line of the given circle and the point-circle. (Certainly, a point circle lying on another circle is necessarily tangent to it.) The equation of the radical line is easily produced by just subtracting the two given equations; thus,
$$(x^2+y^2+6x-4y+3)-(x^2+y^2+4x-10y+29)=0$$
or, after simplification and solving for $y$,
$$y=\frac{13-x}{3}$$
is the sought radical line, which is also the tangent line of the circle $x^2+y^2+6x-4y+3=0$ through $(-2,5)$.
A: For the first part, where did the square root come from? The sign of $3$ is wrong.  
You are right that the tangent line is one that touches the circle in one point.  The slope of the tangent to a curve is given by the derivative at that point.  So rearrange your equation to $y= $ some function of $x$, take $\frac {dy}{dx}$, and evaluate it at $(-2,5)$.  When you rearrange the equation you will have a square root and need to think about the sign.
