Algebraic Tangents

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MaggieThatcher72

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Hello! We had some maths homework over the holidays, and I can't make any progress on the last question. My dad has had a look too, and he's hit a roadblock. Any help would be greatly appreciated.

"A line y=mx+10 is a tangent to the circle x2+y2=25. What are the possible values of m?"
 
Hello, and welcome to FMH! :)

Substitute for \(y\) in the equation of the circle using the equation for the line:

[MATH]x^2+(mx+10)^2=25[/MATH]
Now, you have a quadratic in \(x\) only. If the line is tangent to the circle, the two intersect at only 1 point, what does this tell us about the discriminant of the quadratic in \(x\)?
 
When x= 0, y= m(0)+ 10= 10 so the lines pass through (0, 10). There are two lines that pass through (0, 10) tangent to the circle with center at (0, 0) and radius 5. The three line segments, the segment from (0, 10) to (0, 0), which has length 10, the segment from (0, 0) to the point of tangency, which, as a radius, has length 5, and the tangent line, form a right triangle with one leg of length 5 and hypotenuse of length 10. By the Pythagorean theorem, the other leg, the tangent segment, has length \(\displaystyle \sqrt{100- 25}= \sqrt{75}= 5\sqrt{3}\). So the two tangent segments are tangent to the circle, \(\displaystyle x^2+ y^2= 25\) where that circle intersects \(\displaystyle x^2+ (y- 10)^2= 75\) or \(\displaystyle x^2+ y^2- 20y+ 100- 50= 25\). That will be where \(\displaystyle x^2+ y^2= x^2+ y^2- 20y+ 50\). We can immediately subtract \(\displaystyle x^2+ y^2\) from both sides leaving \(\displaystyle -20y+ 50= 0\) so \(\displaystyle y= 5/2\). Then \(\displaystyle x^2+ \frac{25}{4}= 25\), \(\displaystyle x^2= 25- \frac{25}{4}= \frac{75}{4}\) so \(\displaystyle x= \pm\frac{5}{2}\sqrt{3}\).

The two tangents pass through (0, 10) and \(\displaystyle \left(\frac{5}{2}\sqrt{3}, \frac{5}{2}\right)\) and through (0, 10) and \(\displaystyle \left(-\frac{5}{2}\sqrt{3}, \frac{5}{2}\right)\).
 
MarkFL said:
Hello, and welcome to FMH! :)

Substitute for \(y\) in the equation of the circle using the equation for the line:

[MATH]x^2+(mx+10)^2=25[/MATH]
Now, you have a quadratic in \(x\) only. If the line is tangent to the circle, the two intersect at only 1 point, what does this tell us about the discriminant of the quadratic in \(x\)?
Click to expand...

It tells us that the discriminant is 0. I've got to x2+x2m2+20xm+75=0, but where do I go from here? I can't work out the discriminant until I've consolidated the x2, but how would I do that?
 
Another approach is to use trigonometry. Visualize the objects in the coordinate plane: there is a circle with radius 5 centered on the origin, and a line spinning about the point (0, 10). An interactive version of this can be found here: https://www.desmos.com/calculator/gpgegmujna

Like @HallsofIvy pointed out, if you connect the origin to one of the points where the tangent line intersects the circle, you construct a right triangle. This triangle has a short leg of length 5 (the radius of the circle), a hypotenuse of length 10 (the Y-intercept of the tangent line) and some third length we could calculate but don't care about.

The angle of the vertex at (0, 10) has characteristics such that [MATH]sin^{-1}\left(\frac{5}{10}\right) = 30°[/MATH]. The tangent line, therefore, has an angle of [MATH]\pm60°[/MATH], whose slopes can be calculated from this in a handful of different ways.
 
MaggieThatcher72 said:
It tells us that the discriminant is 0. I've got to x2+x2m2+20xm+75=0, but where do I go from here? I can't work out the discriminant until I've consolidated the x2, but how would I do that?
Click to expand...

Let's write the quadratic in the form:

[MATH](m^2+1)x^2+20mx+75=0[/MATH]
As you correctly stated, the discriminant must be zero:

[MATH](20m)^2-4(m^2+1)(75)=0[/MATH]
Divide by 100:

[MATH](2m)^2-3(m^2+1)=0[/MATH]
Can you proceed?
 
Mark's is the simple solution to determining what m may be. The only change that I might make is here.

[MATH](20m)^2 - 4(m^2 + 1)(75) = 0 \implies 400m^2 - 300m^2 - 300 = 0.[/MATH]
Then I would say to divide by 100.

Hall's answer is really round about. After all his work, he still gets an answer only after this arithmetic:

[MATH]\dfrac{10 - 2.5}{0 - (-\ 2.5\sqrt{3})} = \dfrac{3 * 2.5}{2.5\sqrt{3}} = \dfrac{(\sqrt{3})^2}{\sqrt{3}} = \sqrt{3}.[/MATH]
Mr. Bland's answer may be as simple Mark's if Maggie has learned trigonometry and the trigonometric inverses. Given that she was given pause by how to deal with x^2 + mx^2, I suspect the arcsin function will not assist her.
 
Last edited:
MarkFL said:
Let's write the quadratic in the form:

[MATH](m^2+1)x^2+20mx+75=0[/MATH]
As you correctly stated, the discriminant must be zero:

[MATH](20m)^2-4(m^2+1)(75)=0[/MATH]
Divide by 100:

[MATH](2m)^2-3(m^2+1)=0[/MATH]
Can you proceed?
Click to expand...
Yes thank you! M would be + or - sqrt(3), wouldn't it?
 
Another method we could use is to observe that radius of the circle to the point of tangency will be perpendicular to the tangent line, and so will lie along the line:

[MATH]y=-\frac{1}{m}x[/MATH]
Equating the two lines, we get:

[MATH]mx+10=-\frac{1}{m}x[/MATH]
Solve for \(x\):

[MATH]x=-\frac{10m}{m^2+1}\implies y=\frac{10}{m^2+1}[/MATH]
And the distance from this point to the origin must be 5, so we must have:

[MATH]\left(\frac{10m}{m^2+1}\right)^2+\left(\frac{10}{m^2+1}\right)^2=25[/MATH]
[MATH]10^2(m^2+1)=25(m^2+1)^2[/MATH]
[MATH]4=m^2+1[/MATH]
[MATH]m^2=3[/MATH]
[MATH]m=\pm\sqrt{3}[/MATH]
 
Another method, similar to Mr. Bland's post#5.

This is simple if you're good at Pythagoras and similar triangles (read the left pic first, then the right)...

gradient_of_tangent.png

And obviously there's another solution with m=-√3 on the right of the y axis.
 
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