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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?

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Hint: The radius of a circle is always perpendicular to the tangent. – Giuseppe Negro Mar 1 '13 at 9:36
Can you sketch the tangent line? – oks Mar 1 '13 at 9:54
Please make the title more specific. Right now only one out of the five words convey any information at all about the contents of the question. – Rahul Mar 1 '13 at 10:04

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, $$ \begin{align} \begin{bmatrix} x\\ y \end{bmatrix} &= \frac1{13}\begin{bmatrix} 3&2\\ 2&-3 \end{bmatrix} \begin{bmatrix} 40\\ -21 \end{bmatrix}\\[6pt] &= \begin{bmatrix} 6\\ 11 \end{bmatrix}\\ \end{align} $$

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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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Thanks very much for all the help, I think I've got it now. – doc Mar 1 '13 at 10:33
don't forget to accept an answer :) – Chris Mar 2 '13 at 6:35

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 !

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