Find the equation of the two lines tangent to the circle x^2+y^2-4x+6y-12=0 at x=5

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mcall

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Find the equation of the two lines tangent to the circle x^2+y^2-4x+6y-12=0 at x=5
 
I would begin by completing the square on \(x\) and \(y\), so that you can express the circle in the form:

[MATH](x-h)^2+(y-k)^2=r^2[/MATH]
What do you get?
 
Yes, but let's write it as:

[MATH](x-2)^2+(y+3)^2=5^2[/MATH]
Now, when \(x=5\), what is \(y\)?
 
mcall said:
y=0, right?
Click to expand...

Let's check:

[MATH](5-2)^2+(y+3)^2=5^2[/MATH]
[MATH]3^2+(y+3)^2=5^2[/MATH]
[MATH](y+3)^2=5^2-3^2=4^2[/MATH]
[MATH]y+3=\pm4[/MATH]
[MATH]y=-3\pm4[/MATH]
And so we find:

[MATH]y\in\{-7,1\}[/MATH]
fmh_0057.png

As you can see, I have labeled the two tangent points. Now, to determine the slope of the lines that are tangent to the circle at those points, we can use the fact that the tangent lines will be perpendicular to radii of the circle. We know the center of the circle is at \((2,-3)\).

And so let's do the upper tangent point first. What is the slope of the line through the circle's center and the tangent point?
 
mcall said:
the slope is -3/4
Click to expand...
We can see that -3/4 is not the correct slope because moving 4 units to the right (the 'run') and 3 units down (the 'rise') from the center brings us to the point (6,-6) instead of (5,1). We're talking about the slope of the line which passes through the circle's center (2,-3) and the upper tangent point (5,1). Please show your work, so we can see what you're doing.

Please read the forum's submission guidelines, also. Thanks.

?
 
mcall said:
Find the equation of the two lines tangent to the circle x^2+y^2-4x+6y-12=0 at x=5
Click to expand...
The implicit derivative is:
\(\displaystyle \begin{align*}2x+2yy'-4+6y'&=0 \\(2y+6)y'&=4-2x\\y'&=\frac{4-2x}{2y+6} \end{align*}\)

\(\displaystyle \begin{align*}@(5,1)\quad y'&=\frac{-6}{8}=\frac{-3}{4} \\\\@(5,-7)\quad y'&=\frac{-6}{-8}=\frac{3}{4} \end{align*}\)

@Otis the slope of \(\displaystyle \frac{-3}{4}\quad @(5,1)\) is correct.
 
pka said:
@Otis the slope of \(\displaystyle \frac{-3}{4}\quad @(5,1)\) is correct.
Click to expand...
Thanks. I realized mcall's careless communication in post #7, after I saw post #10.

By the way, your implicit differentiation will be useful going forward, considering all of the calculus students we see on the wrong board.

?
 
I'm going to assume that you found the slope of the line through the two points:

[MATH]\frac{1+3}{5-2}=\frac{4}{3}[/MATH]
And then taking the negative multiplicative inverse of this as the slope of the tangent line, we get:

[MATH]m_1=-\frac{3}{4}[/MATH]
And likewise with the second tangent point, we then find:

[MATH]m_2=\frac{3}{4}[/MATH]
So, you now have the slopes of both tangent lines, and a point on those lines, so to finish it's simply a matter of using the point-slope formula to write the equations of the two tangent lines.

fmh_0058.png

What do you get for the two tangent lines?
 
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