Let $L(h,k)$ be the foot of the perpendicular line segment drawn from $P(10,0)$ on the tangent line. Then the slope of this line segment is given by
$$m=\frac{k}{h-10}.$$
Let the point of tangency be $Q(x_1,y_1)$. Then the equation of the tangent line at $Q$ is given by $xx_1+yy_1=16$. So the slope of this line is $-\frac{x_1}{y_1}$. Using perpendicularity, we get
$$\frac{k}{h-10}=\frac{y_1}{x_1} \implies \color{red}{kx_1+(10-h)y_1=0}.$$
But the point $L$ also lies on it, so
$$\color{red}{hx_1+ky_1=16}.$$
Solving for $x_1$ from the first equation , we get
$$x_1=\frac{(h-10)y_1}{k}.$$
Substituting this into the second equation we get
$$\color{blue}{y_1=\frac{16k}{h(h-10)+k^2}.}$$
Then
$$\color{blue}{x_1=\frac{16(h-10)}{h(h-10)+k^2}.}$$
Now use the fact that $Q(x_1,y_1)$ lies on the circle, to get
$$x_1^2+y_1^2=16 \implies \left[\frac{16(h-10)}{h(h-10)+k^2}\right]^2+\left[\frac{16k}{h(h-10)+k^2}\right]^2=16.$$
This simplifies to
$$(h(h-10)+k^2)^2=16((h-10)^2+k^2).$$
If you consider $\angle ABC = \theta$ in your figure, having $AB=6 \implies CB = 6\cos\theta, \ CA = 6\sin\theta$, and $C$ being the origin, assuming the line moves about in the positive quadrant, we get
$B=(6\cos\theta,0), \ C = (0,6\sin\theta) \implies \text{midpoint of }BC =(3\cos\theta,3\sin\theta), \ 0\le\theta\le \dfrac{\pi}2 $
So the midpoint has co-ordinates $(x,y)=(3\cos\theta,3\sin\theta)$ with $\theta \in \left[0,\dfrac{\pi}2\right] \\ \implies x^2+y^2=9$ but restricted only to the quarter in the first quadrant.
I am adding this as an answer because this way you don't come across the formula you have presented. I tried out some graphs on Desmos by beginning with $$x\sqrt{9-y^2} + y\sqrt{9-x^2}=9 \qquad (1)$$ which, as you have observed doesn't yield a graph, but playing around with the components in this formula, the following do produce graphs $$x\sqrt{9-y^2} + y\sqrt{9-x}=9 \\ x\sqrt{9-y} + y\sqrt{9-x^2}=9 \\ x\sqrt{9-y} + y\sqrt{9-x}=9$$ and the following don't $$\sqrt{9-y^2} + \sqrt{9-x^2}=9 \\ \sqrt{9-y} + \sqrt{9-x}=9 \\ \sqrt{9-y} + \sqrt{9-x^2}=9$$
Coming back to your formula, note that $$x^2+y^2=9 \implies \sqrt{9-y^2}=|x|\ne x$$ so, if Desmos had to graph your formula $(1)$, it should ideally graph both $$x|x|+y|y|=9\text{ and } x^2+y^2=9$$ since squaring adds unnecessary roots as Mick has noted, which have some amount of overlap but these are not entirely the same, so Desmos is kind of confused, I guess, as to which one you want, this definitely has to do with the graphing algorithm.
Point of interest: $$x\sqrt{9-y^2} + y\sqrt{9-x^2}=9$$ leads to both $$x^2+y^2=9 \text{ and } x|x|+y|y|=9$$ and the above two formulae overlap exactly in the quarter of the circle in the first quadrant.
Best Answer
You have not used the information of the length of the rod yet. That gives you $$(x_2-x_1)^2+(0-y_1)^2=4$$
If you plug $x_2=2h+2k, x_1=-2k , y_1=2k$ into this, you will find an equation about $h,k$.