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? 
 A: Let $(x,y)$ be the point of tangency.
The point is on the line:
$$
(x,y)\cdot(3,2)=40
$$
and the line from the center of the circle to the point is perpendicular to the line:
$$
\Big((x,y)-(-6,3)\Big)\cdot(2,-3)=0\\
\Updownarrow\\
(x,y)\cdot(2,-3)=-21
$$
Therefore,
$$
\begin{bmatrix}
3&2\\
2&-3
\end{bmatrix}
\begin{bmatrix}
x\\
y
\end{bmatrix}
=
\begin{bmatrix}
40\\
-21
\end{bmatrix}
$$
Thus,
$$
\begin{align}
\begin{bmatrix}
x\\
y
\end{bmatrix}
&=
\frac1{13}\begin{bmatrix}
3&2\\
2&-3
\end{bmatrix}
\begin{bmatrix}
40\\
-21
\end{bmatrix}\\[6pt]
&=
\begin{bmatrix}
6\\
11
\end{bmatrix}\\
\end{align}
$$
A: Logic - Tangent is perpendicular to radius.
Answer -
Equation of tangent (1) : $3X+2Y = 40  \Rightarrow \text{slope} = -3/2.$
Centre $= (-6,3).$
Let point of contact of tangent on circle $= (x,y)$
Since line joining $(-6,3)$ and $(x,y)$ is radius, it is perpendicular to the tangent => slope of radius = $2/3$   ( $-3/2$ x slope of radius = -1).
Equation of radius (2) with point $(-6,3)$ and slope $2/3 : 2X-3Y = -21.$
Solve (1) and (2) to get the point of contact.
Hope the answer is clear !
A: 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 $(6,11)$.
