Determine equation of circle tangent to both the $x$-axis at $(4, 0)$ and the line $y=x+1$ (and perhaps modify question to be a better exercise)

Question

The context is I am writing for a high school level math competition and was thinking of this interesting question to include. I briefly checked MSE and there's nothing directly similar (most similar questions give you information about the center).

Here's what I have so far:

• Numerically I know the radius satisfies $2.07 < r < 2.08$
• The center should be $(4, r)$ (and so the equation can be written as $(x - 4)^2 + (y - r)^2 = r^2$ and the distance from $(4, r)$ to the other point of tangency $(p, p+1)$ is also $r$. Simplifying this I was able to get that $r = \displaystyle\frac{p^2 - 3p + 8.5}{p + 1}$.
• From here, I don't know what to do. My first thought was to somehow use similar triangles and consider the angle I'm going up from the center to the tangent line point $(p, p+1)$.

EDIT: Using some obvious calculus, I get that $r = 2p - 3$ and so the radius is equal to .

Is there any way to complete this without calculus?

Two questions I have:

1.) Am I on the right track to solving this numerically? If so, what insight am I missing to finish this off?

2.) Can I slightly modify this problem so that the algebra works out nicer for students? I think the obvious choice is to use $y = x$ at the tangent line, but perhaps there is something nicer that doesn't immediately reduce the difficulty that much? Maybe something with 30-60-90 triangles?

Thank you,

Garrett

Answer 1

Since the circle is tangent to the $x$ axis at $(4, 0)$ , then its center is of the form

$ C = (4, y) $

Since the circle is tangent to $ y = x + 1 $, then the distance between $C$ and this line is $| y | $, i.e.

$ r = | y | = \dfrac{ | y - 4 - 1 | }{\sqrt{2}} $

Therefore, we now have

$ \sqrt{2} | y | = | y - 5 | $

Squaring both sides

$ 2 y^2 = y^2 - 10 y + 25 $

So that,

$ y^2 + 10 y - 25 = 0 $

Completing the square,

$ (y +5)^2 = 50 $

The solutions of which are

$y = -5 \pm \sqrt{50} $

These solutions are shown in this Desmos page

Answer 2

The intersection point of the tangents is $(-1,0)$ From this we know the length of tangent is 5 units.

Now we can join this point to the centre $(4,r)$ and form a right triangle. As one of the tangents is the $ x-axis $ itself, a lot of our calculations is simplified.

The other tangent has the slope $1$ which corresponds to $ \tan \frac{\pi}{4} $. So the slope of the angle bisector is $ \tan \frac{\pi}{8} $.

In the right triangle formed by $(-1,0) , (4,0)$ and $(4,r) $, we get

$$ \tan \frac{\pi}{8} = \frac {r}{5} $$ $$ \sqrt {2} -1 = \frac {r}{5} $$ $$ r = 5(\sqrt{2} -1) $$

If this is a high school level math competition then I would assume the students know the trig values. But if you want to make the problem simpler, you can change the slope from $1$ to $\sqrt{3}$ and use a y-intercept which works out nicely with this value. Along with this you could change the point of tangency on x-axis from $4$ to $4\sqrt{3}$.

Hope this helps

EDIT I am quite mad I did not account for all possible cases so I will correct my grave oversight.

Another circle can be made such that it touches $y=x+1$ below the x-axis. Here the angle between the tangents is $\frac{3\pi}{4}$ so the angle bisector is $\frac{3\pi}{8}$.

I am treating r as a distance here. Applying trigonometric relation, we get

$$ \tan \frac{3\pi}{8} = \frac {r}{5} $$ $$ \sqrt {2} +1 = \frac {r}{5} $$ $$ r = 5(\sqrt{2} +1) $$

But the point will below the x-axis so we should add a $-$ sign to this solution.

So finally, $ r=-5(\sqrt{2} +1), 5(\sqrt{2} -1) $