<s>[INST] Given coordinates of center, equation of tangent to a circle, what is their point of contact? A circle centre $(-6,3)$ has a tangent with equation $$3X + 2Y = 40$$ What are the coordinates of the point of contact of the tangent with the circle? 
 [/INST] The radius of the circle must be perpendicular to the tangent. Rewriting $3x+2y=40$ gives us $y=-\frac{3}{2}x+20$, so the slope of the tangent line is $-\frac{3}{2}$. The slope of the perpendicular is the negative reciprocal, so the slope of the radius is $\frac{2}{3}$, and therefore the equation of the line the radius lies on is: $y_r=\frac{2}{3}x+b$. We know the center of the circle $(-6,3)$ is on this line, so we have: $3=\frac{2}{3}(-6)+b\Rightarrow b=7$, so $y_r=\frac{2}{3}x+7$. Now solve the equation $\frac{2}{3}x+7=-\frac{3}{2}x+20$, yielding $x=6$. Plugging $6$ in for $x$ in either equation gives $y=11$, so the coordinates of the point of intersection are $(6,11)$.
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