Given P(-2, 3) and circle (x-3)^2 + (y-3)^2 = 9, find the tangents (of the circle) through P.
A line passing thruogh the point P and havinf the slope m has the equation
$\displaystyle y-3=m(x+2)$
The line must intersect the circle in a single point, so the system of equations formed by the equation of the line and the equation of the circle must have only one solution.
$\displaystyle \left\{\begin{array}{ll}(x-3)^2+(y-3)^2=9\\y-3=m(x+2)\end{array}\right.$
$\displaystyle y=m(x+2)+3$
Replace y in the equation of the circle and you'll find a quadratic equation:
$\displaystyle (1+m^2)x^2+(4m^2-6)x+4m^2=0$
This equation must have equal roots. For this, the discriminant must be 0.
$\displaystyle \Delta=b^2-4ac=-64m^2+36=0\Rightarrow m=\pm\frac{3}{4}$.
Replace m in the equation of the line with these two numbers and you'll find two tangents line to the circle.