Find the equation of the tangent to the circle x^2 + y^2 + 10x + 2y + 13 = 0 at the point (-3, 2)

First, what is a tangent? A line which touches a circle (or ellipse) at just one point. In our case (-3,2)
As it touches the line at one point, it is perpendicular the radius at (-3,2). Which implies that the gradient of the tangent is the negative reciprocal of the gradient of the the radius at (-3,2).
To work out the gradient of the radius at (-3,2) we need to know the centre point of the circle.
Rearrange the equation of the circle into square form yielding, centre as (-5,-1)
With two points on the same line we can work out the gradient of the radius at (-3,2). It's 3/2
So gradient of the tangent is the negative reciprocal of 3/2 which is -2/3.
We now have a point on the tangent line (-3,2) and the gradient of the line, hence we can work out the equation of the tangent. Solving yields 2x+3y=0
NOTE: I will draw a diagram on the whiteboard and do the arithmetic in the session also