GivenP(-2, 3) and circle (x-3)^2 + (y-3)^2 = 9, find the tangents (of the circle) throughP.

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- Jul 15th 2010, 10:26 AMMATNTRNGAnalytical Geometry
Given

*P*(-2, 3) and circle (*x*-3)^2 + (*y*-3)^2 = 9, find the tangents (of the circle) through*P*. - Jul 15th 2010, 12:08 PMred_dog
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.