<s>[INST] 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.
 [/INST] 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
$y=-1/3\cdot x+13/3$ 
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