Circle touching the graph of function y = |x|. Determine the function on which the centers of the circles lie for all values of t

Question

I'm curios if I solved this correctly, or if there are any other ways to do it. The problem goes:

The point P(t, |t|) lies on the graph of the function $f(x) = |x|$. A circle with a radius of $√2·|t|/3$ touches the graph of the function f from above at the point P. Determine the function on which the centers of the circles lie for all values of t.

What I did was express the coordinates of the circle in terms of t. So for any point t on the graph, I find the x coordinate by moving left for $√2*t*cos(45)/3$, and moving up for $√2*t*sin(45)/3$ (this only works for t>0). I got the following:

$x = t - √2*t*cos(45)/3$,

$y = |t|+ √2*t*sin(45)/3$

Then I started plugging in values of t, and found that the function is of all the centers is $y = 2|x|$.

Is this a valid way of solving the problem? It seems a little loose to me.

Answer 1

Your solution is correct. Alternatively, let $C$ the center of the circle. Consider the triangle POC. Since your circle touches the line at P, $O\hat{P}C = 90^{\circ}$.

We know that $\overline{OP} = \sqrt2 |t|$. Also, $\overline{CP}$ is the radius, so $\overline{CP} = \sqrt2 \; |t|/3$. Therefore, $\tan(C\hat{O}P) = 1/3$.

Let $\theta = 45^{\circ} + C\hat{O}P$. We have $\frac{y_C}{x_C} = \frac{\overline{OC} \sin(\theta)}{\overline{OC} \cos(\theta) \;\mathrm{sgn}(t)} = \tan(\theta) \;\mathrm{sgn}(t)$, where $\mathrm{sgn}(t) = \begin{cases}1,&t > 0\\-1&t<0\\\end{cases}$. Of course, we are assuming $t \neq 0$.

However, $\tan(\theta) = \tan(45^{\circ} + \arctan(\frac{1}{3})) = \frac{1 + \frac{1}{3}}{1 - \frac{1}{3}} = 2 \implies \frac{y_C}{x_C} = 2 \;\mathrm{sgn}(t)$, so $C$ lies on the graph of the function $f(x)=|2x|$.